For $k=1,2, \cdots, 2000$, we have
$$\begin{array}{l} \left|y_{k}-y_{k+1}\right|=\left|\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right)-\frac{1}{k+1}\left(x_{1}+x_{2}+\cdots+x_{k+1}\right)\right|= \\ \left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right|= \\ \frac{\left|\left(x_{1}-x_{2}\right)+2\left(x_{2}-x_{3}\right)+\cdots+k\left(x_{k}-x_{k+1}\right)\right|}{k(k+1)} \leqslant \\ \frac{\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|}{k(k+1)} \end{array}$$

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

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

we get
$$\begin{aligned} \sum_{k=F}^{2000=} L y_{k}-y_{k+1} \mid \leqslant & \left|x_{1}-x_{2}\right|\left(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{2000 \times 2001}\right)+ \\ & 2\left|x_{2}-x_{3}\right|\left(\frac{1}{2 \times 3}+\frac{1}{3-4}+\cdots+\frac{1}{2000 \times 2001}\right)+\cdots++ \\ & 2000 \cdot\left|x_{2000}-x_{2001}\right| \cdot \frac{1}{2000 \times 2001}= \\ & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{2001}\right)+\left|-x_{2}-x_{3}\right|\left(1-\frac{2}{2001}\right)+\cdots+ \\ & \left|x_{2000}-x_{200 F}\right|\left(1-\frac{2000}{2001}\right) \leqslant \\ & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{2001}\right)+\left|x_{2}-x_{3}\right|\left(1-\frac{1}{2001}\right)+\cdots+\cdots \\ & \left|x_{2000}-x_{2000}\right|-\left(1-\frac{1}{2001}\right)= \end{aligned}$$

Equality holds if and only if $\left|x_{F}-x_{2}\right|=2001, x_{2}=x_{3}=\cdots=x_{2001}$, specifically by taking $x_{1}=2001$, $x_{2}=x_{3}=\cdots=x_{200 ₹}=0$ the equality can be achieved.

Therefore, the maximum value of $\sum_{k=1}^{2000}-y_{k}=y_{k+1}$ is 2000.