More generally, for odd n,
1n + 2n + 3n + … + nn
is a multiple of n.
To show that, note that nn is a multiple of n and that we can split the sum into pairs of terms (n-k)n + kn summing to a multiple of n.
We have to expand to see that (n-k)n + kn really is divisible by n for odd n:
(n-k)n + kn
= Cn,0 nn - Cn,1 nn-1 k + Cn,2 nn-2 k2 - … + Cn,n-1 n kn-1 - Cn,n kn + kn
= n (Cn,0 nn-1 - Cn,1 nn-2 k + Cn,2 nn-3 k2 - … + Cn,n-1 kn-1)
The last equality is justified because Cn,n = 1, so the last two terms canceled.