There are $2^{11} - 1 = 2047$ non-empty subsets of $\{a_1,a_2,\ldots,a_{11}\}$. If for one such subset $I$ we have $2011 \mid \sum_{a \in I} a$, we are done. If not, by the pigeonhole principle, there must exist two such subsets $I \neq J$ such that $\sum_{a \in I} a \equiv \sum_{b \in J} b \pmod{2011}$. Then $2011 \mid \sum_{a \in I \setminus J} a - \sum_{b \in J \setminus I} b$.