Solution I

Lemma 1: If $k$ is a positive integer, then there exist $0 \le a_1, a_2, \dots, a_{2k} \le 9$ such that $a_1, a_3, \dots, a_{2k-1}$ are odd integers, $a_2, a_4, \dots, a_{2k}$ are even integers, and
$$2^{2k+1} \mid \overline{a_1 a_2 \cdots a_{2k}}$$
Proof of Lemma 1 by mathematical induction.
If $k=1$, it follows from $8 \mid 16$ that the proposition is true.
Assume that if $k = n-1$, the proposition is true.
When $k=n$, let $\overline{a_1 a_2 \cdots a_{2n-2}} = 2^{2n-1}t$ by the inductive hypothesis.
The problem reduces to proving that there exist $1 \le a, b \le 9$ with $a$ odd and $b$ even such that $2^{2n+1} \mid \overline{ab} \times 10^{2n-2} + 2^{2n-1}t$, or $8 \mid \overline{ab} \times 5^{2n-2} + 2t$, or $8 \mid \overline{ab} + 2t$ in view of $5^{2n-2} \equiv 1 \pmod 8$.
It follows from $8|12+4$, $8|14+2$, $8|16+0$ and $8|50+6$ that Lemma 1 is true.

Lemma 2: If $k$ is a positive integer, then there exists an alternating number $\overline{a_1 a_2 \cdots a_{2k}}$ with an even number $2k$ of digits such that $a_{2k}$ is odd and $5^{2k} \mid \overline{a_1 a_2 \cdots a_{2k}}$, where $a_1$ can be 0, but $a_2 \ne 0$.
Proof of Lemma 2 by mathematical induction.
If $k=1$, it follows from $25 \mid 25$ that the proposition is true.
Assume that if $k=n-1$, the proposition is true, or there exists an alternating multiple $\overline{a_1 a_2 \cdots a_{2n-2}}$ satisfying $5^{2n-2} \mid \overline{a_1 a_2 \cdots a_{2n-2}}$.
When $k=n$, let $\overline{a_1 a_2 \cdots a_{2n-2}} = t \cdot 5^{2n-2}$. The problem reduces to proving that there exist $0 \le a, b \le 9$ with $a$ even and $b$ odd such that $5^2 \mid \overline{ab} \times 10^{2n-2} + t \cdot 5^{2n-2}$, or $25 \mid \overline{ab} \times 2^{2n-2} + t$.
Since $2^{2n-2}$ is coprime to 25, there exist $0 < \overline{ab} \le 25$ such that $25 \mid \overline{ab} \times 2^{2n-2} + t$. If $b$ is odd, between $\overline{ab}$ and $\overline{ab} + 50$, at least one satisfies that the highest-valued digit is even. If $b$ is even, between $\overline{ab} + 25$ and $\overline{ab} + 75$, at least one satisfies that the highest-valued digit is even.
This completes the proof of Lemma 2.

Let $n = 2^\alpha 5^\beta t$, where $t$ is coprime to 10 and $\alpha, \beta \in \mathbb{N}$. Assume that $\alpha \ge 2$ and $\beta \ge 1$. Let $l$ be an arbitrary multiple of $n$. The last decimal digit is 0, and the digit in tens is even. Hence these $n$ do not satisfy the required condition.

① When $\alpha = \beta = 0$, consider $21$, $2121$, $212121$, $\dots$, $2121\dots21$, $\dots$. There must exist two of them congruent modulo $n$.
Without loss of generality, we may assume that $t_1 > t_2$, and
$$\underbrace{2121\cdots21}_{\text{number } t_1 \text{ of } 21} \equiv \underbrace{2121\cdots21}_{\text{number } t_2 \text{ of } 21} \pmod{n}.$$
Then
$$\underbrace{2121\cdots21}_{\text{number } t_1-t_2 \text{ of } 21} \equiv \underbrace{00\cdots0}_{\text{number } 2t_2 \text{ of } 0} \equiv 0 \pmod{n}$$
Hence
$$\underbrace{2121\cdots21}_{\text{number } t_1-t_2 \text{ of } 21} \equiv 0 \pmod{n},$$
because $n$ is coprime to 10.
Now these positive integers $n$ satisfy the required condition.

② When $\beta = 0$ and $\alpha \ge 1$, it follows from Lemma 1 that there exists an alternating number $\overline{a_1 a_2 \cdots a_{2k}}$ satisfying $2^\alpha \mid \overline{a_1 a_2 \cdots a_{2k}}$. Consider
$$\overline{a_1 a_2 \cdots a_{2k}}, \overline{a_1 a_2 \cdots a_{2k} a_1 a_2 \cdots a_{2k}}, \dots, \overline{a_1 a_2 \cdots a_{2k} a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}, \dots$$
There must exist two of them congruent modulo $t$. Without loss of generality, we may assume that $t_1 > t_2$, and
$$\overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}} \equiv \overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}} \pmod{t}$$
Since $t$ is coprime to 10, $\overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}} \equiv 0 \pmod{t}$. Moreover, since $t$ is coprime to 2,
$$2^\alpha t \mid \overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}},$$
which is alternating.

③ When $\alpha = 0, \beta \ge 1$, it follows from Lemma 2 that there exists an alternating multiple $\overline{a_1 a_2 \cdots a_{2k}}$ satisfying $5^\beta \mid \overline{a_1 a_2 \cdots a_{2k}}$ with $a_{2k}$ odd. Using the same argument as in ②, we obtain that there exist $t_1 > t_2$ satisfying $t \mid \overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}$. Since $t$ is coprime to 5,
$$5^{\beta_t} \mid \overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}$$
In addition, $\overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}$ is alternating, and the last decimal digit $a_{2k}$ is odd.

④ When $\alpha = 1$ and $\beta \ge 1$, it follows from ③ that there exists an alternating number $\overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}$ satisfying that $a_{2k}$ is odd and $5^{\beta_t} \mid \overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}$. Hence $2 \cdot 5^{\beta_t} \mid \overline{a_1 a_2 \cdots a_{2k} \cdots a_1 a_2 \cdots a_{2k}}$, which is alternating.

In conclusion, if $n$ is not divisible by 20, then these positive integers $n$ satisfy the required condition.

Solution II

$n$ should be the positive integers and should satisfy that $n$ is not divisible by 20.

(1) Assume that $20 \mid n$. Select a multiple of $n$ arbitrarily. Denote $(a_k a_{k-1} \cdots a_1)_{10}$, where $a_k \ne 0$. We have $20 \mid (a_k a_{k-1} \cdots a_1)_{10}$. Hence $a_1 = 0$, $2 \mid (a_k a_{k-1} \cdots a_1)_{10}$. This implies that $a_2$ is even. Therefore, $(a_k a_{k-1} \cdots a_1)_{10}$ is not alternating.

(2) We will prove that if $n$ is a positive integer and is not divisible by 20, then $n$ must have a multiple which is alternating.
We will show three lemmas as follows. Then we divide the proof into four cases.

Lemma 1: If the positive integer $n$ is coprime to 10, then for any $l \in \mathbb{N}$, there exists $k \in \mathbb{N}$ such that
\frac{\begin{array}{c} 1 \ 00\cdots0 \\ \underline{\quad} \\ \mathbf{number} \ l \ \mathbf{of} \ 0 \end{array} \quad \begin{array}{c} 1 \ 00\cdots0 \\ \underline{\quad} \\ \mathbf{number} \ l \ \mathbf{of} \ 0 \end{array} \quad \begin{array}{c} 1 \cdots 1 \ 00\cdots0 \\ \underline{\quad} \\ \mathbf{number} \ l \ \mathbf{of} \ 0 \end{array} \quad \begin{array}{c} 1 \\ \underline{\quad} \\ \mathbf{total number} \ k \ \mathbf{of} \ 1 \end{array}}
is a multiple of $n$.
Proof of Lemma 1: Consider the numbers $x_1 = 1, x_2 = 1 \underbrace{00\cdots0}_{\text{number } l \text{ of } 0}, \dots,$
$$x_m = 1 \underbrace{00\cdots0}_{\text{number } l \text{ of } 0} \ 1 \cdots 1 \underbrace{00\cdots0}_{\text{number } l \text{ of } 0} \ 1, \dots$$
There must exist two of them congruent modulo $n$. Without loss of generality, we may assume that
$$x_s \equiv x_t (\text{mod } n), s > t \ge 1.$$
Then $n \mid x_s - x_t$. Moreover, since $x_s - x_t = x_{s-t} \cdot 10^{t(l+1)}$ and $n$ is coprime to $10$, $n \mid x_{s-t}$, $s-t > 0$ and $x_{s-t}$ is a positive integer. This completes the proof of Lemma 1.

Lemma 2: For any $m \in \mathbb{N}$, there always exists an alternating number $s_m$ with $m$ digits such that its first digit can be 0, and its last decimal digit is 5, and $5^m \mid s_m$.
Proof of Lemma 2 by inductive construction. Start with $s_1 = 5$. Suppose that $s_m = (a_m a_{m-1} \cdots a_1)_{10}$ is alternating, where $a_1 = 5$ and $a_m$ can be 0. In addition, $5^m \mid s_m$. Let $s_m = 5^m$. Denote
$$A = \begin{cases} \{0, 2, 4, 6, 8\}, & \text{when } a_m \text{ is odd,} \\ \{1, 3, 5, 7, 9\}, & \text{when } a_m \text{ is even.} \end{cases}$$
Any two of them in $A$ are not congruent modulo $5$, and $2^m$ is coprime to $5$. So any two of the numbers in $\{2^m x \mid x \in A\}$ are not congruent modulo $5$. Select $x \in A$ such that $2^m x = -l \pmod 5$. Then $5 \mid 2^m x + l$. Let $s_{m+1} = (x a_m a_{m-1} \cdots a_1)_{10}$. Then $s_{m+1}$ is an alternating number with $m+1$ digits, where $a_1 = 5$, and its first digit can be 0. In addition, $s_{m+1} = x 10^m + s_m = 5^m(2^m x + l)$ is a multiple of $5^{m+1}$. This completes the proof of Lemma 2.

Lemma 3: For any integer $m \in \mathbb{N}$, there always exists an alternating number $t_m$ with $m$ digits such that its last decimal digit is 2, and
$$2^{2k+1} \mid t_{2k+1}, \quad 2^{2k+3} \mid t_{2k+2}.$$
Proof of Lemma 3 by inductive construction. Start with $t_1 = 2$. Suppose that $t_{2k+1} = (a_{2k+1} a_{2k} \cdots a_1)_{10}$ is alternating, where $a_1 = 2$, and $2^{2k+1} \mid t_{2k+1}$. It follows from the property of the alternating numbers that $a_{2k+1} = a_1 = 0 \pmod 2$.
Denote $A = \{1, 3, 5, 7\}$. Any two of them in $A$ are not congruent modulo $8$, and $5^{2k+1}$ is coprime to $8$. So any two numbers of $\{5^{2k+1} x \mid x \in A\}$ are not congruent modulo $8$. Now divide the four numbers in $\{5^{2k+1} x \mid x \in A\}$ by $8$ respectively. The original sentence means the 4 numbers are divisible by 8. Since $5^{2k+1}x$ with $x \in A$ is odd, the remainders are 1, 3, 5, 7.
Let $t_{2k+1} = 2^{2k+1}l$ with $l$ odd. Select $x \in A$ such that $5^{2k+1}x \equiv -l+4 \pmod 8$. Then $2^2 \nmid 5^{2k+1}x+l$. Let $t_{2k+2} = (x a_{2k+1} \cdots a_1)_{10}$. Then $t_{2k+2}$ is alternating with $2k+2$ digits. In addition, $a_1 = 2$ and $t_{2k+2} = x 10^{2k+1} + t_{2k+1} = 2^{2k+1}(5^{2k+1}x+l)$.
Since $2^2 \nmid 5^{2k+1}x+l$, we have $2^{2k+3} \nmid t_{2k+2}$.
Moreover, suppose that $t_{2k+2} = (a_{2k+2} \cdots a_1)_{10}$, where $a_1 = 2$, and $2^{2k+3} \nmid t_{2k+2}$. Let $t_{2k+3} = (4 a_{2k+2} \cdots a_1)_{10}$. Then $t_{2k+3}$ is an alternating number with $2k+3$ digits, and $t_{2k+3} = 5^{2k+2}2^{2k+4} + t_{2k+2}$.
Since $2^{2k+3} \nmid t_{2k+2}$, we have $2^{2k+3} \nmid t_{2k+3}$. This completes the proof of Lemma 3.

Next, we discuss the four cases.
(1) If $n$ is coprime to 10, it follows from Lemma 1 that there exists $k \in N^*$ such that $10 \ 101 \cdots 101$ is a multiple of $n$. The conclusion is equivalent to selecting $l=1$ in Lemma 1.
(2) If $n$ is not divisible by 5, and $n$ is divisible by 2, let $n = 2^m n_0$ such that $n_0$ is not divisible by 2, then $n_0$ is coprime to 10. Select $m_0 > m$ with $m_0$ even. It follows from Lemma 3 that there exists an alternating number $t_{m_0} = (a_{m_0} \cdots a_1)_{10}$ with $m_0$ digits, where $a_1 = 2$. In addition, $2^{m_0+1} \nmid t_{m_0}$. Hence $2^m \mid t_{m_0}$. It follows from Lemma 1 that there exists $k \in N^*$ such that
$$P = t_{m_0} \cdot 1 \underbrace{00\cdots0}_{\substack{\text{number } m, -1 \text{ of } 0 \\ \text{number } k \text{ of } 1}} \cdot 1\cdots100\cdots01$$
can be divisible by $2^m n_0$, or $n \mid P$.
(3) If $n$ is divisible by 5, and $n$ is not divisible by 2, let $n = 5^m n_0$ such that $n_0$ is not divisible by 5, then $n_0$ is coprime to 10. Select $m_0 > m$ with $m_0$ even. It follows from Lemma 2 that there exist an alternating number $s_{m_0} = (a_{m_0} \cdots a_1)_{10}$ with $m_0$ digits, where $a_1 = 5$ and $a_{m_0}$ can be 0. In addition, $5^m \mid s_{m_0}$. Hence $5^m \mid s_{m_0}$. Using the same argument as (2), there exists an alternating number $P$ such that $5^m n_0 \mid P$ and the last decimal digit $a_1 = 5$.
(4) If $n$ is divisible by 10, and $n$ is not divisible by 20, let $n = 10n_0$ with $n_0$ odd, then $n_0$ satisfies the assumption in case (1) or in case (3). It follows from (1) and (3) that there exists $(a_k \cdots a_1)_{10}$ an alternating multiple of $n_0$, where $a_1 = 1$ or 5. Now let $P = (a_k \cdots a_1 0)_{10}$. Then $P$ is an alternating multiple of $n$.

In conclusion, $n$ should be a positive integer and should not be divisible by 20.