jmc

Note that $x+2010\equiv x(\mathrm{mod}2010)$. Add 2010 to every negative summand and rearrange the terms to find $S\equiv 0+1+2+3+\cdots +2008+2009(\mathrm{mod}2010)$. The right-hand side is the sum of the integers from 1 to 2010, so $S\equiv \frac{2010\cdot 2011}{2}\equiv 1005\cdot 2011\equiv 1005\cdot 1\equiv \boxed{1005}(\mathrm{mod}2010)$.