Let $s=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be a permutation of the natural numbers $1,2, \cdots, n$ over the past $n$ months. $f(s)$ is defined as the minimum value of the absolute difference between any two adjacent elements in $s$. Find the maximum value of $f(s)$.
Solution: We discuss in two cases.
(1) If $n=2 k$ is even, then $f(s) \leqslant k$ and the absolute difference between it and its adjacent numbers is no more than $k$. On the other hand, in
$$
\begin{array}{l}
s=(k+1,1, k+2,2, \cdots, 2 k, k) \text {, we have } \\
f(s)=k=\left[\frac{n}{2}\right] .
\end{array}
$$
(2) If $n=2 k+1$ is odd, then $f(s) \leqslant k+1$ and the absolute difference between it and its adjacent numbers is no more than $k$. On the other hand, in
$$
s=(k+1,1, k+2,2, \cdots, 2 k, k, 2 k+1)
$$

we have
$$
f(s)=k=\left[\frac{n}{2}\right] .
$$

In summary, $\max f(s)=\left[\frac{n}{2}\right]$.