4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups.
4. 12 .
$$
\begin{array}{r}
\text { Since }(a+1)(b+1) \\
=2009=41 \times 7^{2} \text {, and given }
\end{array}
$$

that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps.