Slovenian Mathematical Olympiad, 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 $i \neq j$. 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 $n$.