If
$$
f(x)=\sum_{i=1}^{n} N_{i}\left|x-a_{i}\right|\left(N_{i} \in \mathbf{N}_{+}, a_{1}<a_{2}<\cdots<a_{n}\right) \text {, }
$$

then the minimum value of $f(x)$ is attained at some segment point in the middle or between two adjacent segment points.
Solve: From the problem, we have
$$
\begin{array}{l}
f(x)=\sum_{k=1}^{2011} k\left|x-\frac{1}{k}\right| \\
=|x-1|+\left(\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{2}\right|\right)+\cdots+ \\
(\left\lvert\, \underbrace{\left|x-\frac{1}{2011}\right|+\cdots+\left|x-\frac{1}{2011}\right|}_{\text {2011 }} .\right.
\end{array}
$$

The total number of terms is
$$
1+2+\cdots+2011=1006 \times 2011 \text {. }
$$

According to the arrangement pattern, the corresponding extremum point is $x=\frac{1}{m}$, where the positive integer $m$ satisfies
$$
\begin{array}{l}
\frac{(m-1) m}{2}<\frac{1006 \times 2011}{2} \leqslant \frac{m(m+1)}{2} \\
\Rightarrow m=1422 \\
\Rightarrow f(x)_{\min }=f\left(\frac{1}{1422}\right)=\frac{1}{1422} \sum_{k=1}^{2011}|k-1422| \\
\quad=\frac{1}{1422}\left(\frac{1422 \times 1421}{2}+\frac{589 \times 590}{2}\right) \\
\quad=\frac{592043}{711} .
\end{array}
$$