10. (20 points) Find
$$
f(x)=|x-1|+2|x-2|+\cdots+2011|x-2011|
$$

the minimum value.
10. Obviously, when $x=2011$, $f(x)$ has no minimum value.
The following assumes $x \in[1,2011]$.
When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2010)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2011} i(i-x) \\
=\left(k^{2}+k-2011 \times 1006\right) x+ \\
\quad \frac{2011 \times 2012 \times 4023}{6}- \\
\quad \frac{k(k+1)(2 k+1)}{3}
\end{array}
$$

is a linear function, and its minimum value is attained at the endpoints of the interval $[k, k+1]$.

Therefore, the minimum term of the sequence $\{f(k)\}(1 \leqslant k \leqslant 2010)$ is the desired value.
Notice that,
$$
\begin{array}{l}
f(k+1)>f(k) \\
\Leftrightarrow k^{2}+k-2011 \times 1006>0 \\
\Leftrightarrow k \geqslant 1422 . \\
\text { Hence } f(x)_{\min }=f(1422) \\
=1422\left(1422^{2}+1422-2011 \times 1006\right)+ \\
\frac{2011 \times 2012 \times 4023}{6}-\frac{1422 \times 1423 \times 2845}{3} \\
=794598996 .
\end{array}
$$