At first, we consider the case that $n = 2^r$, $r \ge 1$. Obviously, at this time $r < k$. We take three $2^{r-1}$'s and $2k-3$ $1$'s — each of them cannot be divided by $n$. If these $2k$ numbers are divided into two groups, then there must exist a group that contains two $2^{r-1}$'s, whose sum is $2^r$ — divisible by $n$.

Next, we consider the case that $n$ is not a power of $2$. At this time, the $2k$ integers we take are
$$-1, -1, -2, -2^2, \dots, -2^{k-2}, 1, 2, 2^2, \dots, 2^{k-1}.$$
Then they each cannot be divided by $n$.
Assume these numbers can be divided into two groups, such that any partial sum of numbers in one group is not divisible by $n$. We may say that $1$ is in the first group. Since $(-1) + 1 = 0$ is divisible by $n$, the two $-1$'s must be in the second group; since $(-1) + (-1) + 2 = 0$, the $2$ is in the first group; then the $-2$ is in the second group.

Now by induction, assuming $1, 2, \dots, 2^l$ are in the first group and $-1, -2, -2^2, \dots, -2^l$ in the second one ($1 \le l < k-2$), since
$$(-1) + (-1) + (-2) + \dots + (-2^l) + 2^{l+1} = 0$$
is divisible by $n$, we get that $2^{l+1}$ is in the first group, and then $-2^{l+1}$ in the second.

Therefore, $1, 2, 2^2, \dots, 2^{k-2}$ is in the first group and $-1, -2, -2^2, \dots, -2^{k-2}$ in the second. Finally, since
$$(-1) + (-1) + (-2) + \dots + (-2^{k-2}) + 2^{k-1} = 0,$$
then $2^{k-1}$ is in the first group. Therefore, $1, 2, 2^2, \dots, 2^{k-1}$ are all in the first group.

On the other hand, the knowledge about the binary number system tells us that every positive integer which is not greater than $2^k - 1$ can be represented as the partial sum of $1, 2, 2^2, \dots, 2^{k-1}$. Since $n \le 2^k - 1$, then it is the partial sum of $1, 2, 2^2, \dots, 2^{k-1}$ that is of course divisible by $n$ itself. This is a contradiction to the assumption.

Therefore, we have found out $2k$ integers that meet the requirement in the question. The proof is then complete. ☐