Problem 945

[Solution]
$$\begin{array}{l} \left|y_{k}-y_{k+1}\right|=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\ = \left.\frac{1}{k(k+1)} \right\rvert\,\left(x_{1}-x_{k+1}\right)+\left(x_{2}-x_{k+1}\right)+\cdots+\left(x_{k}-\right. \\ \left.x_{k+1}\right) \mid \\ \leqslant \frac{1}{k(k+1)}\left(\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k \mid x_{k}\right. \\ -\frac{\left.x_{k+1} \mid\right)}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\right|, \end{array}$$

Thus, $\left|y_{1}-y_{2}\right|+\left|y_{2}-y_{3}\right|+\cdots+\left|y_{1992}-y_{1993}\right|$
$$\begin{aligned} & =\sum_{k=1}^{1992}\left|y_{k}-y_{k+1}\right| \\ = & \sum_{k=1}^{1992} \sum_{i=1}^{k} \frac{i}{k(k+1)}\left|x_{i}-x_{i+1}\right| \\ = & \sum_{i=1}^{1992} \sum_{k=1}^{1992} \frac{i}{k(k+1)}\left|x_{i}-x_{i+1}\right| \\ = & \left.\sum_{i=1}^{1992} i\left[\frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\cdots+\frac{1}{1992 \cdot 1993}\right] \cdot \right\rvert\, x_{i}- \\ & x_{i+1}^{1992} \mid \\ \leqslant & \sum_{i=1}^{1992}\left(\frac{1}{i}-\frac{1}{1993}\right) \cdot\left|x_{i}-x_{i+1}\right| \\ = & \left(1-\frac{1}{1993}\right) \cdot\left|x_{i}-x_{i+1}\right| \\ = & \left(1-\frac{1}{1993}\right) \cdot \sum_{i=1}^{1992}\left|x_{i}-x_{i+1}\right| \\ = & 1992 . \end{aligned}$$

On the other hand, if we let $x_{1}=t+1993, x_{2}=x_{3}=\cdots=x_{1993}=t$,
then
$$\begin{array}{l} \left.y_{k}-y_{k+1}\left|=\frac{i}{k(k+1)} \cdot\right| x_{1}-x_{2} \right\rvert\,=\frac{1993}{k(k+1)} \\ \sum_{k=1}^{1992}\left|y_{k}-y_{k+1}\right|=1993 \cdot \sum_{k=1}^{1992} \frac{1}{k(k+1)}=1992 . \end{array}$$

Therefore, the maximum value of $\sum_{k=1}^{1992}\left|y_{k}-y_{k+1}\right|$ is 1992.