n even, easy problem

I think you're right. $n$ must indeed be even, because otherwise $1+2+\ldots+n=\frac{n(n+1)}2$ is divisible by $n$. Conversely, if $n$ is even, then we can arrange the numbers $2,1,4,3,6,5,\ldots,2k,2k-1$. It should be fairly easy to check that this arrangement satisfies the condition (not sure; just wrote it in a hurry ).

yes it does, prove it by induction as an exercise.