5. The number of positive integers $n$ such that $n+1$ divides $n^{2006}+2006$ is $\qquad$.

Makes the positive integer $n$ such that $n+1$ can divide $n^{2006}+2006$ total $\qquad$.

Note: The second sentence seems to be a repetition or a different phrasing of the first. If it's meant to be a different statement, please clarify. Otherwise, I will assume the first translation is sufficient.
5. 5 .

From the problem, we have
$$
\begin{array}{l}
n^{2006}+2006 \equiv(-1)^{2006}+2006=2007 \\
\equiv 0(\bmod (n+1)) .
\end{array}
$$

Since $2007=3 \times 3 \times 223$, then
$$
n+1=3,9,223,669,2007 \text {. }
$$

Therefore, $n=2,8,222,668,2006$.