51st Ukrainian National Mathematical Olympiad, 3rd Round

Let $a_1$ be the first number and $a_n$ is the last one. We arrive at: $$\frac{a_1+a_n}{2}n=2011\iff (a_1+a_n)n=2\cdot 2011.$$ Since $2$ and $2011$ are prime, we have two options: $a_1+a_n=2$ and $n=2011$ or $a_1+a_n=2011$ and $n=2$. First case is clearly impossible, while the second case gives the answer: $1005+1006$.