Proving $\sum_{k=1}^m{k^n}$ is divisible by $\sum_{k=1}^m{k}$ for $ n=2013$ I got an interesting new question, it's about number theory and algebra precalculus. Here is the question:

a positive integer $n$ is called valid if $1^n+2^n+3^n+\dots+m^n$ is divisible by $1+2+3+\dots+m$ for every positive integer $m$.
  
  
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*Prove that 2013 is valid
  
*Prove that there are infinite positive integers which are not valid
  

Every little hint, contribution and recommendation would be very helpful. Sorry for my bad english. Thanks before.

If $n$ is odd, then modulo $m+1$ we have $2(1^n + 2^n + \ldots + m^n) = (1^n+m^n) + (2^n+(m-1)^n) + \ldots + (m^n+1^n) \\ \equiv (1^n-1^n) + (2^n-2^n) + \ldots + (m^n-m^n) = 0 \pmod {m+1}$.
Also, since $m^n \equiv 0 \pmod m$, $2(1^n + 2^n + \ldots + m^n) \equiv 2(1^n + 2^n + \ldots + (m-1)^n) \equiv 0 \pmod m$
Since $m$ and $m+1$ are coprime, this shows that $2(1^n + \ldots + m^n)$ is a multiple of $m(m+1)$, and since $m(m+1)$ is even, $1^n + \ldots + m^n$ is a multiple of $m(m+1)/2 = 1+2+\ldots+m$.
If $n$ is even then $1+2^n \equiv 2 \pmod 3$, which is not a multiple of $1+2 = 3$
