13. If the function $f(x)=\ln \frac{\mathrm{e} x}{\mathrm{e}-x}$, then $\sum_{k=1}^{2010} f\left(\frac{k e}{2011}\right)=$ $\qquad$ .
13.2010.

Notice
$$
\begin{array}{l}
f(x)+f(\mathrm{e}-x) \\
=\ln \left[\frac{\mathrm{e} x}{\mathrm{e}-x} \cdot \frac{\mathrm{e}(\mathrm{e}-x)}{\mathrm{e}-(\mathrm{e}-x)}\right]=2 . \\
\text { Therefore } \sum_{k=1}^{2010} f\left(\frac{k \mathrm{e}}{2011}\right) \\
=\sum_{k=1}^{1005}\left(f\left(\frac{k \mathrm{e}}{2011}\right)+f\left(\mathrm{e}-\frac{k \mathrm{e}}{2011}\right)\right) \\
=2 \times 1005=2010 .
\end{array}
$$