Example 6 Real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy $\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001$, let $y_{k}$ $=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2, \cdots, 2001$, find the maximum possible value of $\sum_{k=1}^{2000} \mid y_{k}-y_{k+1}$. (2001 Shanghai High School Mathematics Competition Question)
For $k=1,2, \cdots, 2000$, we have
$$\begin{aligned}
& \quad\left|y_{k}-y_{k+1}\right| \\
& =\left|\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right)-\frac{1}{k+1}\left(x_{1}+x_{2}+\cdots+x_{k+1}\right)\right| \\
& \quad=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right| \\
& =\frac{\left|\left(x_{1}-x_{2}\right)+2\left(x_{2}-x_{3}\right)+\cdots+k\left(x_{k}-x_{k+1}\right)\right|}{k(k+1)} \\
& \quad \leqslant \frac{\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|}{k(k+1)}
\end{aligned}$$

By the identity
$$\begin{aligned}
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{(n-1) n} \\
& =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)=1-\frac{1}{n}
\end{aligned}$$

and its result
$$\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\cdots+\frac{1}{(n-1) n}=\frac{1}{k}\left(1-\frac{k}{n}\right)$$

we get
$$\begin{aligned}
& \sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \\
\leqslant & \left|x_{1}-x_{2}\right|\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{2000 \cdot 2001}\right) \\
& +2\left|x_{2}-x_{3}\right|\left(\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{2000 \cdot 2001}\right) \\
& +\cdots+2000 \cdot\left|x_{2000}-x_{2001}\right| \cdot \frac{1}{2000 \cdot 2001} \\
= & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{2001}\right)+\left|x_{2}-x_{3}\right|\left(1-\frac{2}{2001}\right)+\cdots \\
& +\left|x_{2000}-x_{2001}\right|\left(1-\frac{2000}{2001}\right) \\
\leqslant & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{2001}\right)+\left|x_{2}-x_{3}\right|\left(1-\frac{1}{2001}\right)+\cdots \\
& +\left|x_{2000}-x_{2001}\right|\left(1-\frac{1}{2001}\right) \\
= & \left(1-\frac{1}{2001}\right) \sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2000
\end{aligned}$$

Equality holds if and only if $\left|x_{1}-x_{2}\right|=2001, x_{2}=x_{3}=\cdots=x_{2001}$. In particular, taking $x_{1}=2001, x_{2}=x_{3}=\cdots=x_{2001}=0$ makes the equality hold. Therefore, the maximum value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$ is 2000.