9・50 For every permutation $S$ of $(1,2, \cdots, n)$ where $n \geqslant 2$, let $f(S)$ be the minimum value of the absolute differences between every two adjacent elements in $S$. Find the maximum value of $f(S)$.
[Solution] Let $n=2k$ be an even number. For any $s$, since the absolute difference between each element $k$ and its adjacent elements is $\leqslant k$, we have
$$
f(s) \leqslant k.
$$

On the other hand, take $s=(k+1,1, k+2,2, \cdots, 2k, k)$, then $f(s)=k$, thus we get
$$
\max f(s)=k=\left[\frac{n}{2}\right].
$$

Let $n=2k+1$, for any $s$, since the absolute difference between $k+1$ and its adjacent elements is $\leqslant k$, thus $f(s) \leqslant k$. If we take
$$
S=(k+1,1, k+2,2, \cdots, 2k, k, 2k+1),
$$

then $f(s)=k$. Thus $\max f(s)=k=\left[\frac{n}{2}\right]$.
In any case, we have $\max f(s)=\left[\frac{n}{2}\right]$.