For which $k$, there is no integer pair $(x,y)$ such that $x^2 - y^2 = k$?

$ 
\textbf{(A)}\ 2005
\qquad\textbf{(B)}\ 2006
\qquad\textbf{(C)}\ 2007
\qquad\textbf{(D)}\ 2008
\qquad\textbf{(E)}\ 2009
$
1. We start with the given equation \(x^2 - y^2 = k\). This can be factored as:
   \[
   x^2 - y^2 = (x-y)(x+y) = k
   \]
2. We need to determine for which values of \(k\) there are no integer pairs \((x, y)\) that satisfy this equation.

3. Consider the factorization \(k = (x-y)(x+y)\). For \(k\) to be expressed as a product of two integers, both \(x-y\) and \(x+y\) must be integers.

4. If \(k\) is even, then both \(x-y\) and \(x+y\) must be either both even or both odd. This is because the product of two even numbers or two odd numbers is even.

5. If \(k\) is odd, then one of \(x-y\) or \(x+y\) must be even and the other must be odd. This is because the product of an even number and an odd number is odd.

6. Now, let's analyze the given options:
   - \(2005\) is odd.
   - \(2006\) is even.
   - \(2007\) is odd.
   - \(2008\) is even.
   - \(2009\) is odd.

7. For \(k\) to be even, we need to check if \(k\) can be written as a product of two integers where both are either even or both are odd. Specifically, if \(k\) is even, it must be divisible by 4 for it to be written as a product of two integers where both are even or both are odd.

8. Let's check the even numbers:
   - \(2006\) is even but not divisible by 4 (since \(2006 \mod 4 = 2\)).
   - \(2008\) is even and divisible by 4 (since \(2008 \mod 4 = 0\)).

9. Since \(2006\) is even but not divisible by 4, it cannot be written as a product of two integers where both are even or both are odd. Therefore, there are no integer pairs \((x, y)\) such that \(x^2 - y^2 = 2006\).

Conclusion:
\[
\boxed{2006}
\]