16. If the integer $m$ makes the equation
$$
x^{2}-m x+m+2006=0
$$

have non-zero integer roots, then the number of such integers $m$ is
$\qquad$.
16.5.

Let the two integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=m, a \beta=m+2006, \\
\text { i.e., } \alpha \beta-(\alpha+\beta)+1=2006+1 \\
=2007=(\alpha-1)(\beta-1) .
\end{array}
$$

Thus, $\alpha-1= \pm 1, \pm 3, \pm 9$;
$$
\beta-1= \pm 2007, \pm 669, \pm 223 \text {. }
$$

Therefore, there are 5 such integers $m$.