solved here

Gryphos wrote: My solution: Let $f(x)=x^4-x^3-x+1=(x-1)(x^3-1)=(x-1)^2(x^2+x+1)$, then we have $f(x) \geq 0$ for all real $x$ with equality only for $x=1$. But from the two equations we easily get $\sum_{k=1}^{2000} f(x_k) = 0$. So we must have $x_1=x_2=\cdots=x_{2000}=1$, which is indeed a solution.