8. Given that $k$ is a positive integer not exceeding 2008, such that the equation $x^{2}-x-k=0$ has two integer roots. Then the sum of all such positive integers $k$ is $\qquad$ .
8.30360

Since the roots of $x^{2}-x-k=0$ are $x=\frac{1 \pm \sqrt{4 k+1}}{2}$, therefore,
$x^{2}-x-k=0$ has two integer roots $\Leftrightarrow 4 k+1$ is a square of an odd number.
Let $1+4 k=(2 a+1)^{2}\left(a \in \mathbf{N}_{+}\right)$. Then $k=a(a+1)$.
Also, $k \leqslant 2008$, so $a \leqslant 44$.
Therefore, the sum of all such $k$ is
$$
\begin{array}{l}
1 \times 2+2 \times 3+\cdots+44 \times 45 \\
= \frac{1}{3}(1 \times 2 \times 3-0 \times 1 \times 2)+ \\
\frac{1}{3}(2 \times 3 \times 4-1 \times 2 \times 3)+\cdots+ \\
\frac{1}{3}(44 \times 45 \times 46-43 \times 44 \times 45) \\
= \frac{1}{3} \times 44 \times 45 \times 46=30360 .
\end{array}
$$