55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

From the first equation $-1\le x_i\le 1$, $i=1,\dots ,2015$, so for all $i$: $0\le 1+x_i\le 2$. Add both equations and get $$x_1^{2014}(1+x_1)+x_2^{2014}(1+x_2)+\dots +x_{2015}^{2014}(1+x_{2015})=0.$$ Since every item $x_i^{2014}(1+x_i)\ge 0$, then their sum equals zero if and only if every item equals zero. So all $x_i\in \{-1,0\}$. From the first equation follows that exactly one variable is not null, from the second equation follows that this variable equals $-1$. Thus, the equation is satisfied only by such set of numbers: $x_i=-1$, $i=1,\dots ,2015$, $x_j=0$, $j\ne i$.