If $n$ and $m$ are positive integers, and $n$ is even while $m$ is odd, then the fraction \[\frac {1^m+2^m+3^m+\dots+n^m} {1+2+3+\dots+n}\] is always integer. Indeed, let $n=2k$. Then the denominator is $\frac {2k(2k+1)} 2=k(2k+1)$. The numbers $k$ and $2k+1$ are coprime. We can represent the numerator in two different ways: \[(1^m+(2k-1)^m)+(2^m+(2k-2)^m)+(3^m+(2k-3)^m)+\dots+\\+((k-1)^m+(k+1)^m)+k^m+(2k)^m\] and \[(1^m+(2k)^m)+(2^m+(2k-1)^m)+(3^m+(2k-2)^m)+\dots+\\+(k^m+(k+1)^m)\] and then use the known fact that $x^m+y^m$ is divisible by $x+y$ when $m$ is odd.