$6 \cdot 90$ Let the numbers $x_{1}, \cdots, x_{1991}$ satisfy the condition
$$
\left|x_{1}-x_{2}\right|+\cdots+\left|x_{1990}-x_{1991}\right|=1991 \text {, }
$$

and $y_{k}=\frac{1}{k}\left(x_{1}+\cdots+x_{k}\right), k=1, \cdots, 1991$. Try to find the maximum value that the following expression can achieve.
$$
\left|y_{1}-y_{2}\right|+\cdots+\left|y_{1990}-y_{1991}\right| .
$$
[Solution] For $k=1, \cdots, 1990$, we have the relation
$$
\begin{aligned}
& \left|y_{k}-y_{k+1}\right| \\
= & \left|\frac{1}{k}\left(x_{1}+\cdots+x_{k}\right)-\frac{1}{k+1}\left(x_{1}+\cdots+x_{k+1}\right)\right| \\
= & \left|\frac{1}{k(k+1)}\left(x_{1}+\cdots+x_{k}-k x_{k+1}\right)\right| \\
\leqslant & \frac{1}{k(k+1)}\left[\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|\right],
\end{aligned}
$$

From this, we get
$$
\begin{aligned}
& \left|y_{1}-y_{2}\right|+\cdots+\left|y_{1990}-y_{1991}\right| \\
\leqslant & \left|x_{1}-x_{2}\right|\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{1990 \cdot 1991}\right) \\
& +2\left|x_{2}-x_{3}\right|\left(\frac{1}{2 \cdot 3}+\cdots+\frac{1}{1990 \cdot 1991}\right) \\
& +\cdots+1990\left|x_{1990}-x_{1991}\right| \frac{1}{1990 \cdot 1991} \\
= & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{1991}\right)+\left|x_{2}-x_{3}\right|\left(1-\frac{2}{1991}\right)+\cdots+ \\
& \left|x_{1990}-x_{1991}\right|\left(1-\frac{1990}{1991}\right) \\
\leqslant & 1991 \cdot\left(1-\frac{1}{1991}\right)=1990 .
\end{aligned}
$$

The obtained estimate can be achieved when $x_{1}=1991, x_{2}=\cdots=x_{1991}=0$, so the maximum value sought is 1990.