(10) Let $[x]$ denote the greatest integer not exceeding $x$, then the value of $\sum_{k=1}^{2008}\left[\frac{2008 k}{2009}\right]$ is $\qquad$
For $k=1,2, \cdots, 2008$, since $\frac{2008 k}{2009}$ is not an integer, we have
$$
\begin{aligned}
& {\left[\frac{2008 k}{2009}\right]+\left[\frac{2008(2009-k)}{2009}\right] } \\
= & {\left[\frac{2008 k}{2009}\right]+\left[2008-\frac{2008 k}{2009}\right] } \\
= & 2007,
\end{aligned}
$$

Thus, we have
$$
\begin{aligned}
& \sum_{k=1}^{2008}\left[\frac{2008 k}{2009}\right] \\
= & \frac{1}{2} \sum_{k=1}^{2008}\left(\left[\frac{2008 k}{2009}\right]+\left[\frac{2008(2009-k)}{2009}\right]\right) \\
= & \frac{1}{2} \times 2008 \times 2007 \\
= & 2015028 .
\end{aligned} .
$$