China Mathematical Competition (Complementary Test)

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 $$(-1)+(-1)+(-2)+\cdots +(-2^l)+2^{l+1}=0$$ 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 $$(-1)+(-1)+(-2)+\cdots +(-2^{k-2})+2^{k-1}=0,$$ 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. ☐