Any idea?

Here is my idea, not complete though. Let $P(x)=a_{2n-1}x^{2n-1}+\cdots+a_1x + a_0 = \sum_{k=0}^{2n-1}a_kx^k$. Now observe that, $$ x^n\equiv 1 \pmod{x^{n-1}+\cdots+1} \implies P(x)\equiv (a_{2n-1}+a_{n-1})x^{n-1}+(a_{2n-2}+a_{n-2})x^{n-2}+\cdots+(a_n+a_0) \pmod{x^{n-1}+\cdots+1}. $$Hence, letting $b_k=a_k+a_{n+k}$ for $k=0,\dots,n-1$; you are looking at a polynomial, $R(x)=\sum_{k=0}^{n-1}b_k x^k$ and demanding when $x^{n-1}+\cdots+1\mid R(x)$. This happens, if and only if, $$ a_{2n-1}+a_{n-1}=a_{2n-2}+a_{n-2}=\cdots=a_n+a_0. $$Letting $f(k)$ to be the number of ordered pairs $(x,y)$ with $x+y=k$, $1\leq x,y\leq m$; we are to count, $$ \sum_{k=2}^{2m}f(k)^n. $$

Thank you very much,for this nice idea!!