Slovenija 2008

If $k=1$, each of the numbers $a_1,a_2,\dots ,a_n$ is divisible by $n$, so $n$ also divides their sum.

Now, let $1<k<n$ and let $i\ne j$. Since the set $\{a_1,a_2,\dots ,a_n\}\setminus \{a_i,a_j\}$ has $n-2\ge k-1$ elements, one can choose an arbitrary subset $S$ with $k-1$ elements. Then $S\cup \{a_i\}$ and $S\cup \{a_j\}$ are two $k$-tuples of numbers and their sums are divisible by $n$, so the difference of the two sums must also be divisible by $n$. This difference is equal to $a_i-a_j$, so $n$ divides $a_i-a_j$. Here $i$ and $j$ were chosen arbitrarily, so $a_1,a_2,\dots ,a_n$ give the same remainder when divided by $n$, and there are $n$ of them, so their sum must be divisible by $n$.