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
$$\begin{array}{l} 1 \times 2+2 \times 3+\cdots+44 \times 45 \\ = \frac{1}{3}(1 \times 2 \times 3-0 \times 1 \times 2)+ \\ \frac{1}{3}(2 \times 3 \times 4-1 \times 2 \times 3)+\cdots+ \\ \frac{1}{3}(44 \times 45 \times 46-43 \times 44 \times 45) \\ = \frac{1}{3} \times 44 \times 45 \times 46=30360 . \end{array}$$