Positive integers $ a$, $ b$, $ c$, and $ d$ satisfy $ a > b > c > d$, $ a \plus{} b \plus{} c \plus{} d \equal{} 2010$, and $ a^2 \minus{} b^2 \plus{} c^2 \minus{} d^2 \equal{} 2010$. Find the number of possible values of $ a$.
1. Given the equations:
   \[
   a + b + c + d = 2010
   \]
   \[
   a^2 - b^2 + c^2 - d^2 = 2010
   \]

2. We can rewrite the second equation using the difference of squares:
   \[
   a^2 - b^2 + c^2 - d^2 = (a+b)(a-b) + (c+d)(c-d)
   \]

3. Since \(a > b > c > d\) and all are positive integers, we can assume \(a - b \geq 1\) and \(c - d \geq 1\). To satisfy the equality \(a^2 - b^2 + c^2 - d^2 = 2010\), we need to consider the simplest case where \(a - b = 1\) and \(c - d = 1\). This is because any larger difference would make the sum exceed 2010.

4. If \(a - b = 1\) and \(c - d = 1\), then:
   \[
   b = a - 1 \quad \text{and} \quad d = c - 1
   \]

5. Substitute these into the sum equation:
   \[
   a + (a-1) + c + (c-1) = 2010
   \]
   Simplify:
   \[
   2a + 2c - 2 = 2010
   \]
   \[
   2a + 2c = 2012
   \]
   \[
   a + c = 1006
   \]

6. Substitute \(b = a - 1\) and \(d = c - 1\) into the difference of squares equation:
   \[
   a^2 - (a-1)^2 + c^2 - (c-1)^2 = 2010
   \]
   Simplify:
   \[
   a^2 - (a^2 - 2a + 1) + c^2 - (c^2 - 2c + 1) = 2010
   \]
   \[
   a^2 - a^2 + 2a - 1 + c^2 - c^2 + 2c - 1 = 2010
   \]
   \[
   2a + 2c - 2 = 2010
   \]
   \[
   2a + 2c = 2012
   \]
   \[
   a + c = 1006
   \]

7. Since \(a + c = 1006\) and \(a > c\), the possible values for \(a\) range from \(504\) to \(1004\) (since \(a\) and \(c\) are positive integers and \(a > c\)).

8. Therefore, the number of possible values for \(a\) is:
   \[
   1004 - 504 + 1 = 501
   \]

The final answer is \(\boxed{501}\).