### Instruction: Let $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$. What is the residue of $S$, modulo 2010?### Response: I notice that the sum is alternating signs of consecutive integers, starting from 1 and ending at 2010.
This means that I can group the terms in pairs, such as $(1 - 2) + (3 - 4) + \cdots + (2009 - 2010)$.
Each pair has a sum of -1, so there are 1005 pairs in total, and the sum of all the pairs is $-1005$.
To find the residue of $S$, modulo 2010, I need to find the smallest non-negative integer that is congruent to $S$ modulo 2010.
Since $S = -1005$, I can add 2010 to it until I get a positive number.
Adding 2010 once gives $S + 2010 = 1005$.
This is already a positive number, and it is the smallest one that is congruent to $S$ modulo 2010.
Therefore, the residue of $S$, modulo 2010, is 1005.