51st Ukrainian National Mathematical Olympiad, 4th Round

Since $2011$ is a prime number, every multiple of the left-hand side must be equal to either $1$ or $2011$. If $x\ge 0$, the first multiple equals $3x$, and there are no solutions. Similarly, there are no solutions if $y\le 0$ (the second multiple is $-3y$). Suppose now that $x<0$ and $y>0$. Then $$|x+|x+|x||\cdot ||-y|-y|-y|=|x+|x-x||\cdot ||y-y|-y|=|x|\cdot |-y|=-xy=-2011.$$ Recalling that $2011$ is prime, we get the above answers.