Each of the $2N = 2004$ real numbers $x_1, x_2, \ldots , x_{2004}$ equals either $\sqrt 2 -1 $ or $\sqrt 2 +1$. Can the sum $\sum_{k=1}^N x_{2k-1}x_2k$ take the value $2004$? Which integral values can this sum take?
1. Given \(2N = 2004\), we have \(N = 1002\). Each \(x_i\) is either \(\sqrt{2} - 1\) or \(\sqrt{2} + 1\).

2. Let \(x = \sqrt{2} + 1\). Then, we have:
   \[
   x + \frac{1}{x} = 2\sqrt{2}
   \]
   Squaring both sides:
   \[
   x^2 + 2 + \frac{1}{x^2} = 8 \implies x^2 + \frac{1}{x^2} = 6
   \]

3. Since \(x_i \in \{\sqrt{2} - 1, \sqrt{2} + 1\}\), we have:
   \[
   x_i x_{i+1} \in \{x^2, \frac{1}{x^2}, 1\}
   \]

4. We need to find the sum:
   \[
   \sum_{k=1}^{N} x_{2k-1} x_{2k}
   \]
   Let \(\lambda\) be the number of pairs \((x_{2k-1}, x_{2k})\) such that \(x_{2k-1} x_{2k} = x^2\), and \(\mu\) be the number of pairs such that \(x_{2k-1} x_{2k} = \frac{1}{x^2}\). The remaining pairs will have \(x_{2k-1} x_{2k} = 1\).

5. Therefore, we can write:
   \[
   \sum_{k=1}^{N} x_{2k-1} x_{2k} = \lambda x^2 + \mu \frac{1}{x^2} + (N - \lambda - \mu)
   \]

6. Given that the sum is 2004, we have:
   \[
   \lambda x^2 + \mu \frac{1}{x^2} + (N - \lambda - \mu) = 2004
   \]

7. Substituting \(x^2 = 6\) and \(\frac{1}{x^2} = \frac{1}{6}\), we get:
   \[
   6\lambda + \frac{\mu}{6} + (1002 - \lambda - \mu) = 2004
   \]

8. Simplifying the equation:
   \[
   6\lambda + \frac{\mu}{6} + 1002 - \lambda - \mu = 2004
   \]
   \[
   5\lambda + 1002 + \frac{\mu}{6} - \mu = 2004
   \]
   \[
   5\lambda + 1002 - \frac{5\mu}{6} = 2004
   \]
   \[
   5\lambda - \frac{5\mu}{6} = 1002
   \]
   \[
   30\lambda - 5\mu = 6012
   \]
   \[
   6\lambda - \mu = 1202.4
   \]

9. Since \(\lambda\) and \(\mu\) must be integers, the above equation does not hold. Therefore, the sum cannot be 2004.

10. Now, we need to find the possible integral values of the sum. We know:
    \[
    6\lambda + \frac{\mu}{6} + (1002 - \lambda - \mu)
    \]
    Simplifying:
    \[
    5\lambda + 1002 - \frac{5\mu}{6}
    \]

11. Since \(\lambda\) and \(\mu\) are integers, the possible values of \(\lambda\) range from 0 to 501 (since \(N/2 = 501\)).

12. Therefore, the possible values of the sum are:
    \[
    4\lambda + 1002
    \]
    where \(\lambda \in \{0, 1, 2, \ldots, 501\}\).

13. Thus, the possible integral values of the sum are:
    \[
    \{1002, 1006, 1010, \ldots, 3006\}
    \]

The final answer is \( \boxed{ \{1002, 1006, 1010, \ldots, 3006\} } \).