For how many integers $a$ with $|a| \leq 2005$, does the system
$x^2=y+a$
$y^2=x+a$ 
have integer solutions?
1. **Consider the system of equations:**
   \[
   x^2 = y + a
   \]
   \[
   y^2 = x + a
   \]

2. **First, consider the case when \( x = y \):**
   \[
   x^2 = x + a
   \]
   Rearrange to form a quadratic equation:
   \[
   x^2 - x - a = 0
   \]
   Solve for \( x \) using the quadratic formula:
   \[
   x = \frac{1 \pm \sqrt{1 + 4a}}{2}
   \]
   For \( x \) to be an integer, \( 1 + 4a \) must be a perfect square. Let \( 1 + 4a = k^2 \) for some integer \( k \):
   \[
   4a = k^2 - 1
   \]
   \[
   a = \frac{k^2 - 1}{4}
   \]
   For \( a \) to be an integer, \( k^2 - 1 \) must be divisible by 4. Since \( k^2 \equiv 1 \pmod{4} \) for odd \( k \), \( k \) must be odd.

3. **Next, consider the case when \( y = -x - 1 \):**
   \[
   x^2 = -x - 1 + a
   \]
   \[
   x^2 + x + 1 = a
   \]
   Substitute \( y = -x - 1 \) into the second equation:
   \[
   (-x - 1)^2 = x + a
   \]
   \[
   x^2 + 2x + 1 = x + a
   \]
   \[
   x^2 + x + 1 = a
   \]
   This is the same equation as derived above.

4. **Determine the range of \( a \):**
   Given \( |a| \leq 2005 \), we have:
   \[
   -2005 \leq a \leq 2005
   \]
   \[
   -2005 \leq \frac{k^2 - 1}{4} \leq 2005
   \]
   Multiply through by 4:
   \[
   -8020 \leq k^2 - 1 \leq 8020
   \]
   \[
   -8019 \leq k^2 \leq 8021
   \]
   Since \( k^2 \) must be non-negative:
   \[
   0 \leq k^2 \leq 8021
   \]
   The largest integer \( k \) such that \( k^2 \leq 8021 \) is \( k = 89 \) (since \( 89^2 = 7921 \) and \( 90^2 = 8100 \)).

5. **Count the number of valid \( k \):**
   Since \( k \) must be odd, we consider the odd integers from \(-89\) to \(89\):
   \[
   -89, -87, -85, \ldots, -1, 1, 3, \ldots, 87, 89
   \]
   This is an arithmetic sequence with the first term \( a_1 = -89 \), common difference \( d = 2 \), and last term \( a_n = 89 \). The number of terms \( n \) in this sequence is given by:
   \[
   a_n = a_1 + (n-1)d
   \]
   \[
   89 = -89 + (n-1) \cdot 2
   \]
   \[
   89 = -89 + 2n - 2
   \]
   \[
   2n = 180
   \]
   \[
   n = 90
   \]

The final answer is \(\boxed{90}\).