33. Let $\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001, y_{k}=\frac{1}{k} \sum_{i=1}^{k} x_{i}, k=1,2, \cdots, 2001$. Find $\max \sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$
33. Let $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots, 2000$, then $x_{1}=a_{0}, x_{k} = \sum_{i=0}^{k-1} a_{i}, k=2,3, \cdots, 2001$. Thus, the condition becomes $\sum_{k=1}^{2000}\left|a_{k}\right|=2001$, and $y_{k} = \frac{1}{k}\left(a_{0}+\left(a_{0}+a_{1}\right)+\cdots+\left(a_{0}+\cdots+a_{k-1}\right)\right) = \frac{1}{k}\left(k a_{0}+(k-1) a_{1}+\cdots+2 a_{k-2}+a_{k-1}\right), y_{k+1} = \frac{1}{k+1} .\left((k+1) a_{0}+k a_{1}+\cdots+2 a_{k-1}+a_{k}\right)$, therefore, $\left|y_{k}-y_{k+1}\right| = \frac{1}{k(k+1)}\left|a_{1}+2 a_{2}+\cdots+k a_{k}\right| \leqslant \frac{1}{k(k+1)}\left(\left|a_{1}\right|+2\left|a_{2}\right|+\cdots + k\left|a_{k}\right|\right)$. Thus, $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \leqslant \sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \sum_{k=1}^{2000} k\left|a_{k}\right| \leqslant \sum_{k=1}^{2000}\left|a_{k}\right| - \frac{1}{2001} \sum_{k=1}^{2000}\left|a_{k}\right| = \frac{2000}{2001} \sum_{k=1}^{2000}\left|a_{k}\right| = 2000$, equality holds if and only if $\left|a_{1}\right|=2001, a_{2}=a_{3}=\cdots =a_{2000}=0$.