Example 4 Real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy $\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001$, let $y_{k}=\frac{1}{k} \sum_{i=1}^{k} x_{i}, k=1,2, \cdots, 2001$, find the maximum possible value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$.
Let $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots, 2000$, then $x_{1}=a_{0}, x_{k}=\sum_{i=0}^{k-1} a_{i}, k=2,3$, $\cdots, 2001$, the given conditions are transformed into $\sum_{k=1}^{2000}\left|a_{k}\right|=2001$, at this point
$$
\begin{array}{l}
y_{k}=\frac{1}{k}\left[a_{0}+\left(a_{0}+a_{1}\right)+\cdots+\sum_{i=1}^{k-1} a_{i}\right]=\frac{1}{k}\left[k a_{0}+(k-1) a_{1}+\cdots+a_{k-1}\right] \\
y_{k+1}=\frac{1}{k+1}\left[(k+1) a_{0}+k a_{1}+\cdots+2 a_{k-1}+a_{k}\right] .
\end{array}
$$

Therefore, $\left|y_{k}-y_{k+1}\right|=\left|\frac{1}{k(k+1)}\left(-a_{1}-2 a_{2}-\cdots-k a_{k}\right)\right| \leqslant \frac{1}{k(k+1)}\left(\left|a_{1}\right|+2 \mid a_{2}\right.$ $\left.|+\cdots+k| a_{k} \mid\right)$.

Let $A_{k}=\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+k\left|a_{k}\right|, k=1,2, \cdots, 2000, A_{0}=0, a_{2001}=0$, then by the transformation of Abel's summation formula, we have
$$
\begin{aligned}
& \sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \leqslant \sum_{k=1}^{2000}\left(\frac{1}{k}-\frac{1}{k+1}\right) A_{k} \\
= & \sum_{k=1}^{2000} \frac{1}{k}\left(A_{k}-A_{k-1}\right)-\frac{1}{2001} \cdot A_{2000}=\sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \cdot A_{2000},
\end{aligned}
$$

Also, $A_{2000}=\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+2000\left|a_{2000}\right| \geqslant \sum_{k=1}^{2000}\left|a_{k}\right|$, hence
$$
\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \leqslant \sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \cdot \sum_{k=1}^{2000}\left|a_{k}\right|=\frac{2000}{2001} \cdot \sum_{k=1}^{2000}\left|a_{k}\right|=2000 .
$$

From the solution process, we know that the equality holds if and only if $\left|a_{1}\right|=2001, a_{2}=a_{3}=\cdots=a_{2000}=0$, thus the maximum value sought is 2000.