An algebra puzzle

$n$ is odd or even.

If it's odd, then $(n-1)/2$ is an integer. It's easy to see that for every $0<k<(n-1)/2+1$ integer $n|k^n+(n-k)^n$ . (Because $(-k)^n=-k^n$ .) We know that $n|n^n$ , so $n|1^n+(n-1)^n+2^n+(n-2)^n+...=1^n+2^n+...+(n-1)^n.$

If n is even, then let $x$ be the biggest integer such that $2^x$ is a divisor of $n$ . Since $2^x>x$ , if $k$ is an even positive integer then $2^x|k^n$ , if $k$ is odd then from Euler's theorem, we get $k^{2^x-1}\equiv 1$ $(\mod 2^x),$ so $k^n \equiv 1$ $(\mod 2^x)$ . Now we get $1^n+3^n+5^n+...+(n-1)^n \equiv (n/2)$ $(\mod 2^x)$ . From $2^n+4^n+...+(n-2)^n \equiv 0$ $(\mod 2^x), 1^n+2^n+3^n+...+(n-1)^n \equiv (n/2)$ $(\mod 2^x)$ . But if $n|1^n+2^n+3^n+...+(n-1)^n$ then $2^x|(n/2)$ and $2^{x+1}|n$ , which is not possible.

So the only solutions are the odd postive integers. The number of odd positive integers under $2017$ is $1008$ .