Problem 945

$$\text { 17. } \begin{aligned} \left|y_{k}-y_{k+1}\right| & =\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\ & =\frac{1}{k(k+1)}\left|\left(x_{1}-x_{k+1}\right)+\left(x_{2}-x_{k-1}\right)+\cdots+\left(x_{k}-x_{k+1}\right)\right| \\ & \leqslant \frac{1}{k(k+1)}\left(\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|\right) \\ & =\frac{1}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\right| \end{aligned}$$

Thus, $\left|y_{1}-y_{2}\right|+\left|y_{2}-y_{3}\right|+\cdots+\left|y_{1992}-y_{1993}\right|=\sum_{k=1}^{1992}\left|y_{k}-y_{k+1}\right|$
$$\begin{array}{l} =\sum_{k=1}^{1992} \sum_{i=1}^{k} \frac{i}{k(k+1)}\left|x_{i}-x_{i+1}\right|=\sum_{i=1}^{19921992} \sum_{k=1}^{1} \frac{i}{k(k+1)}\left|x_{i}-x_{i+1}\right| \\ =\sum_{i=1}^{1992} i\left[\frac{1}{i(i+1)}+\frac{1}{(i+1)(i+2)}+\cdots+\frac{1}{1992 \cdot 1993)}\right] \cdot\left|x_{i}-x_{i+1}\right| \\ =\sum_{i=1}^{1992} i\left(\frac{1}{i}-\frac{1}{1993}\right) \cdot\left|x_{i}-x_{i+1}\right| \leqslant \sum_{i=1}^{1992}\left(1-\frac{1}{1993}\right) \cdot\left|x_{i}-x_{i+1}\right| \\ =\left(1-\frac{1}{1993}\right) \cdot \sum_{i=1}^{1,992}\left|x_{i}-x_{i+1}\right|=\left(1-\frac{1}{1993}\right) \cdot 1993=1992 . \end{array}$$

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

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