Let $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$. What is the residue of $S$, modulo 2010?
Note that $x + 2010 \equiv x \pmod{2010}$. Add 2010 to every negative summand and rearrange the terms to find $S \equiv 0 + 1 + 2 + 3 + \cdots + 2008 + 2009 \pmod{2010}$. The right-hand side is the sum of the integers from 1 to 2010, so $S \equiv \frac{2010 \cdot 2011}{2} \equiv 1005 \cdot 2011 \equiv 1005 \cdot 1 \equiv \boxed{1005} \pmod{2010}$.
1005
Level 4
Number Theory
Problem: Let $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$. What is the residue of $S$, modulo 2010?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the residue of the sum $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$ modulo 2010, we first recognize the pattern in the series. The series alternates between positive and negative integers. 

We can group the terms in pairs: 
$(1 - 2) + (3 - 4) + (5 - 6) + \cdots + (2009 - 2010)$. Each pair simplifies to $-1$. 

Next, we determine how many such pairs exist in the series. The total number of terms from 1 to 2010 is 2010, which is even. Therefore, the number of pairs is $2010 / 2 = 1005$.

Calculating the sum of these pairs gives:
$S = 1005 \times (-1) = -1005$.

Now, we need to find the residue of $S$ modulo 2010. To do this, we compute:
$-1005 \mod 2010$.

Since $-1005$ is negative, we can convert it to a positive equivalent by adding 2010:
$-1005 + 2010 = 1005$.

Thus, the residue of $S$ modulo 2010 is $1005$.<|END_OF_LATENT|> The final answer is $1005$.