Try to prove that for some $n-k$ we have at least four different ways, but for a bigger number than this $n-k$ we can't find(just look for a counterexample) Also use this well-known lemma: If we have $n$ numbers $a_{1},a_{2},a_{3}\cdots a_{n}$, we can also find a group of numbers from these such that their sum is divisible by $n$ proofLet us denote $S_{1}=a_{1}$ ; $S_{2}=a_{1}+a_{2}$ ; $\cdots$ $S_{n}=\sum_{i=1}^{n} a_{i}$. Now, if one of this sums is divisible by n, then we already have a group from $a_{1},a_{2},\cdots a{n}$ with the sum divisible by $n$. If none of them is, from the Pigeon hole principle we have at least two sums $S_{i}\equiv S_{j} (mod {n})$ so their difference(let $i$ be greater than $j$) is: $S_{i}-S_{j}=a_{j+1}+a_{j+2}+\cdots +a_{i}$ is a sum divisible by $n$. So we're done