For how many integers $1\leq n \leq 2010$, $2010$ divides $1^2-2^2+3^2-4^2+\dots+(2n-1)^2-(2n)^2$?

$ \textbf{(A)}\ 9
\qquad\textbf{(B)}\ 8
\qquad\textbf{(C)}\ 7
\qquad\textbf{(D)}\ 6
\qquad\textbf{(E)}\ 5
$
1. We start by simplifying the given expression:
   \[
   1^2 - 2^2 + 3^2 - 4^2 + \dots + (2n-1)^2 - (2n)^2
   \]
   We can pair the terms as follows:
   \[
   (1^2 - 2^2) + (3^2 - 4^2) + \dots + ((2n-1)^2 - (2n)^2)
   \]
   Each pair can be simplified using the difference of squares formula:
   \[
   a^2 - b^2 = (a - b)(a + b)
   \]
   For each pair \((2k-1)^2 - (2k)^2\), we have:
   \[
   (2k-1)^2 - (2k)^2 = (2k-1 - 2k)(2k-1 + 2k) = -1 \cdot (4k-1) = -(4k-1)
   \]
   Summing these results for \(k\) from 1 to \(n\), we get:
   \[
   -(4 \cdot 1 - 1) - (4 \cdot 2 - 1) - \dots - (4n - 1)
   \]
   This simplifies to:
   \[
   -(4 \cdot 1 + 4 \cdot 2 + \dots + 4n - n) = -4(1 + 2 + \dots + n) + n
   \]
   Using the formula for the sum of the first \(n\) integers:
   \[
   1 + 2 + \dots + n = \frac{n(n+1)}{2}
   \]
   We get:
   \[
   -4 \left( \frac{n(n+1)}{2} \right) + n = -2n(n+1) + n = -2n^2 - 2n + n = -2n^2 - n
   \]

2. We need to find the number of integers \(1 \leq n \leq 2010\) such that \(2010\) divides \(-2n^2 - n\). This is equivalent to finding \(n\) such that:
   \[
   2010 \mid 2n^2 + n
   \]
   Since \(2010 = 2 \times 3 \times 5 \times 67\), we need \(2n^2 + n\) to be divisible by each of these prime factors.

3. Consider the divisibility conditions:
   - For \(2\), \(2n^2 + n\) is always even, so this condition is satisfied for all \(n\).
   - For \(3\), \(2n^2 + n \equiv 0 \pmod{3}\). This simplifies to:
     \[
     2n^2 + n \equiv 0 \pmod{3}
     \]
     This can be factored as:
     \[
     n(2n + 1) \equiv 0 \pmod{3}
     \]
     So, either \(n \equiv 0 \pmod{3}\) or \(2n + 1 \equiv 0 \pmod{3}\), which gives \(n \equiv 1 \pmod{3}\).

   - For \(5\), \(2n^2 + n \equiv 0 \pmod{5}\). This simplifies to:
     \[
     n(2n + 1) \equiv 0 \pmod{5}
     \]
     So, either \(n \equiv 0 \pmod{5}\) or \(2n + 1 \equiv 0 \pmod{5}\), which gives \(n \equiv 2 \pmod{5}\).

   - For \(67\), \(2n^2 + n \equiv 0 \pmod{67}\). This simplifies to:
     \[
     n(2n + 1) \equiv 0 \pmod{67}
     \]
     So, either \(n \equiv 0 \pmod{67}\) or \(2n + 1 \equiv 0 \pmod{67}\), which gives \(n \equiv 33 \pmod{67}\).

4. Using the Chinese Remainder Theorem (CRT), we combine these congruences. For each prime factor \(p\), there are two possible congruences for \(n\). Therefore, there are \(2^3 = 8\) possible values of \(n\) that satisfy all the conditions.

The final answer is \(\boxed{8}\)