Let $n = p_1^{a_1} \cdots p_k^{a_k}$ be the standard factorization of $n$. Since $p_1^{a_1}, \dots, p_k^{a_k}$ are pairwise coprime, by the Chinese Remainder Theorem, for each $i$, $1 \le i \le k$, congruence equations
$$\begin{cases} x \equiv 1 \pmod{p_i^{a_i}} \\ x \equiv 0 \pmod{p_j^{a_j}}, \quad j \neq i \end{cases}$$
have solution $x_i$.
For any solution of $x_0^2 = x_0 \pmod{n}$, we see that $x_0(x_0 - 1) \equiv 0 \pmod{n}$. Then for each $i = 1, 2, \dots, k$, either $x_0 \equiv 0 \pmod{p_i^{a_i}}$ or $x_0 \equiv 1 \pmod{p_i^{a_i}}$. Further, let $S(A)$ be the sum of elements of subset $\{x_1, x_2, \dots, x_k\}$ (particularly, $S(\emptyset) = 0$). Obviously, we have
$$S(A)(S(A) - 1) \equiv 0 \pmod{n}.$$
(This is because of the selection of $x_i$, such that $S(A) \pmod{p_i^{a_i}}$ is either 0 or 1.) Moreover if $A \neq A'$, then $S(A) \neq S(A') \pmod{n}$. Therefore, the sum of all subsets of $\{x_1, x_2, \dots, x_n\}$ is exactly all solutions of $x(x - 1) \equiv 0 \pmod{n}$.
Let $S_0 = n$, $S_r$ be the least non-negative remainder of $x_1 + x_2 + \dots + x_r$ modulo $n$, $r = 1, 2, \dots, k$. Thus $S_k = 1$. For all $1 \le r \le k - 1$, $S_r \neq 0$. Since $k+1$ numbers $S_0, S_1, \dots, S_k$ are in $[1, n]$, by Dirichlet's Drawer Principle, there exist $0 \le l < m \le k$, such that $S_l, S_m$ in the same interval $\left(\frac{jn}{k}, \frac{(j+1)n}{k}\right]$, $(0 \le j \le k-1)$, where $l=0$ and $m=k$ do not hold simultaneously.
Thus, $|S_l - S_m| < \frac{n}{k}$. Denote $y_1 = S_1, y_r = S_r - S_{r-1}$ ($r=2, 3, \dots, k$). So any sum of $y_r \equiv x_r \pmod{n}$ ($r=1, 2, \dots, k$) meets the requirement.
If $S_m - S_l > 1$, then $a = y_{l+1} + y_{l+2} + \dots + y_m = S_m - S_l \in (1, \frac{n}{k})$ is the solution of the equation $x^2 - x \equiv 0 \pmod{n}$.
If $S_m - S_l = 1$, then $n$ ($y_1 + y_2 + \dots + y_l + (y_{m+1} + y_{m+2} + \dots + y_k)$, that is, $n$ ($x_1 + x_2 + \dots + x_l + (x_{m+1} + x_{m+2} + \dots + x_k)$. Notice that $m > l$, which contradicts to the definition of $x_i$.
If $S_m - S_l = 0$, then $n \mid y_{l+1} + y_{l+2} + \cdots + y_m$, that is, $n \mid x_{l+1} + x_{l+2} + \cdots + x_m$, which contradicts the definition of $x_i$.
If $S_m - S_l < 0$, then
$$\begin{aligned} a &= (y_1 + y_2 + \cdots + y_l) + (y_{m+1} + \cdots + y_k) \\ &= S_k - (S_m - S_l) = 1 - (S_m - S_l) \end{aligned}$$
is the solution of equation $x^2 - x \equiv 0 \pmod{n}$, and $1 < a < 1 + \frac{n}{k}$.
Summing up, there exists $a$ satisfying the condition. $\square$