$$\begin{array}{l} \text { III. Solution: From } \left.\left|b_{k}-b_{k+1}\right|=\frac{1}{k(k+1)} \right\rvert\,(k+1)\left(a_{1}+a_{2}\right. \\ \left.+\cdots+a_{k}\right)-k\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \mid \\ =\frac{1}{k(k+1)}\left|a_{1}+a_{2}+\cdots+a_{k}-k a_{k+1}\right| \\ \left.=\frac{1}{k(k+1)} \right\rvert\,\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)+3\left(a_{3}-a_{4}\right)+ \\ \cdots+=k\left(a_{k}-a_{k+1}\right) \mid . \end{array}$$

We have $\left|b_{1}-b_{2}\right|+\left|b_{2}-b_{3}\right|+\cdots+\left|b_{1996}-b_{1997}\right|$
$$\begin{aligned} & =\frac{1}{1 \times 2}\left|a_{1}-a_{2}\right|+\frac{1}{2 \times 3}\left|\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)\right| \\ & \quad+\frac{1}{3 \times 4}\left|\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)+3\left(a_{3}-a_{4}\right)\right| \\ & \left.\quad+\cdots+\frac{1}{1996 \times 1997} \right\rvert\,\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right) \\ & +\cdots+1996\left(a_{1996}-a_{1997}\right) \mid \\ & \leqslant\left(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{1996 \times 1997}\right)\left|a_{1}-a_{2}\right| \\ & +2\left(\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{1996 \times 1997}\right)\left|a_{2}-a_{3}\right| \\ & \quad+\cdots+1996 \frac{1}{1996 \times 1997}\left|a_{1996}-a_{1997}\right| \\ = & \left(1-\frac{1}{1997}\left|a_{1}-a_{2}\right|+\left(1-\frac{2}{1997}\right)\left|a_{2}-a_{3}\right|+\cdots\right. \\ + & \left(1-\frac{1996}{1997}\right)\left|a_{1996}-a_{1997}\right| \\ = & \left(\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{1996}-a_{1997}\right|\right) \\ & -\frac{1}{1997}\left(\left|a_{1}-a_{2}\right|+2\left|a_{2}-a_{3}\right|+\cdots\right. \\ & \left.+1996\left|a_{1996}-a_{1997}\right|\right) \end{aligned}$$
$\leqslant 1997-\frac{1}{1997}\left(\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\mid a_{1996}\right.$
$-a_{1997} \mid$ )
$$=1997-1=1996 \text {. }$$

When $a_{1}=1997, a_{2}=a_{3}=\cdots=a_{1997}=0$, the equality holds in the above expression.
Therefore, the maximum value is 1996.
$$$$