Example 4 Real numbers $x_{1}, x_{2}, \cdots, x_{2} 0001$ satisfy
$$
\begin{array}{c}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001, \\
\text { let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2, \cdots,
\end{array}
$$
2001 . Find the maximum possible value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$.
(2001, Shanghai High School Mathematics Competition)
Explanation: Since the relationship between $x_{k}$ and $x_{k+1}$ is unknown, the difference $x_{k}-x_{k+1}$ can be treated as a whole, and the condition $\sum_{k=1}^{2000} |x_{k}-x_{k+1}|=2001$ can be viewed as a constraint on $x_{k}-x_{k+1}$. By making the substitution $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots, 2000$, we have $x_{1}=a_{0}, x_{k}=\sum_{i=0}^{k-1} a_{i}, k=2,3, \cdots, 2001$, and the condition becomes $\sum_{k=1}^{2000} |a_{k}|=2001$. At this point,
$$
\begin{array}{l}
y_{k}=\frac{1}{k}\left[a_{0}+\left(a_{0}+a_{1}\right)+\cdots+\sum_{i=0}^{k-1} a_{i}\right] \\
=\frac{1}{k}\left[k a_{0}+(k-1) a_{1}+\cdots+a_{k-1}\right], \\
y_{k+1}=\frac{1}{k+1}\left[(k+1) a_{0}+k a_{1}+\cdots+2 a_{k-1}+a_{k}\right].
\end{array}
$$

Then $\left|y_{k}-y_{k+1}\right|$
$$
\begin{array}{l}
=\frac{1}{k(k+1)}\left|-a_{1}-2 a_{2}-\cdots-k a_{k}\right| \\
\leqslant \frac{1}{k(k+1)}\left(\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+k\left|a_{k}\right|\right).
\end{array}
$$

Let $A_{k}=\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+k\left|a_{k}\right|, k=1,2, \cdots, 2000, A_{0}=0$, then
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \leqslant \sum_{k=1}^{2000}\left(\frac{1}{k}-\frac{1}{k+1}\right) A_{k} \\
=\sum_{k=1}^{2000} \frac{1}{k}\left(A_{k}-A_{k-1}\right)-\frac{1}{2001} \cdot A_{2000} \\
=\sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \cdot A_{2000}.
\end{array}
$$
Also, $A_{2000}=\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+2000\left|a_{2000}\right| \geqslant \sum_{k=1}^{2000}\left|a_{k}\right|$, so
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \\
\leqslant \sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \cdot \sum_{k=1}^{2000}\left|a_{k}\right| \\
=\frac{2000}{2001} \cdot \sum_{k=1}^{2000}\left|a_{k}\right|=2000.
\end{array}
$$

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