Four, (18 points) The real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001 . \\
\text { Let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2 .
\end{array}
$$
$\cdots, 2$ 001. Find the maximum possible value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$.
For $k=1,2, \cdots, 2000$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right| \\
=\frac{\left|\left(x_{1}-x_{2}\right)+2\left(x_{2}-x_{3}\right)+\cdots+k\left(x_{k}-x_{k+1}\right)\right|}{k(k+1)} \\
\leqslant \frac{\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|}{k(k+1)} .
\end{array}
$$

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

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

we can obtain $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$
$$
\begin{aligned}
\leqslant & \left|x_{1}-x_{2}\right|\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{2000 \cdot 2001}\right) \\
& +2\left|x_{2}-x_{3}\right|\left(\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{2000 \cdot 2001}\right) \\
& \quad+\cdots+2000 \cdot\left|x_{2000}-x_{2001}\right| \cdot \frac{1}{2000 \cdot 2001} \\
= & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{2001}\right)+\left|x_{2}-x_{3}\right|\left(1-\frac{2}{2001}\right) \\
& \quad+\cdots+\left|x_{20000}-x_{2001}\right|\left(1-\frac{2000}{2001}\right) \\
\leqslant & \sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|\left(1-\frac{1}{2001}\right)=2000 .
\end{aligned}
$$

Equality holds if and only if $\left|x_{1}-x_{2}\right|=2001, x_{2}=x_{3}=\cdots=x_{2001}$, especially when $x_{1}=2001, x_{2}=x_{3}=\cdots=x_{2001}=0$, the equality can be achieved.
Therefore, the maximum value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$ is 2000.
(Proposed by: Li Dayuan, Liu Hongkun, Xiong Bin, Ye Shengyang, Yu Yinglong)