Let $a_{1}, a_{2}, \cdots, a_{n}$ represent any permutation of the integers 1, 2, $\cdots, n$. Let $f(n)$ be the number of such permutations that satisfy: (1) $a_{1}=1$; (2) $\left|a_{i}-a_{i+1}\right| \leqslant 2, i=1$, $2, \cdots, n-1$. Determine whether $f(1996)$ is divisible by 3.
Solution: It is easy to find that $f(1)=f(2)=1, f(3)=2$.
When $n \geqslant 4$, we have $a_{1}=1, a_{2}=2$ or 3.
(a) When $a_{2}=2$, we proceed as follows: delete $a_{1}$, $b_{1}=a_{2}-1, b_{2}=a_{3}-1, \cdots, b_{n-1}=a_{n}-1$, to get a permutation $b_{1}, \cdots, b_{n-1}$ that meets the conditions, and the number of such permutations is $f(n-1)$.
(b) When $a_{2}=3$, for $a_{3}$ we have two cases:
( $\left.1^{\circ}\right) a_{3}=2$, then $a_{4}=4$. By the same process as (a), the number of permutations that meet the conditions is $f(n-3)$;
$\left(2^{\circ}\right) a_{3} \neq 2$, then 2 must be after 4. This leads to the conclusion that the odd numbers are arranged in reverse order, i.e., $1357 \cdots 8642$, which is one permutation.
Therefore, $f(n)=f(n-1)+f(n-3)+1$.
For $n \geqslant 1$, let $r(n)$ be the remainder when $f(n)$ is divided by 3, then we have
$$
\begin{array}{l}
r(1)=r(2)=1, r(3)=2, \\
r(n)=r(n-1)+r(n-3)+1 .
\end{array}
$$

Thus, $r(n+8)=r(n+7)+r(n+5)+1$
$$
\begin{aligned}
= & {[r(n+6)+r(n+4)+1] } \\
& +[r(n+4)+r(n+2)+1]+1 \\
= & r(n+6)+2 r(n+4)+r(n+2)+3 \\
= & {[r(n+5)+r(n+3)+1] } \\
& +2 r(n+4)+r(n+2)+3 \\
= & \{[(r(n+4)+r(n+2)+1] \\
& +[r(n+2)+r(n)+1]+1\} \\
& +2 r(n+4)+r(n+2)+3 \\
= & 3 r(n+4)+3 r(n+2)+6+r(n),
\end{aligned}
$$

which means $r(n+8)$ has the same remainder as $r(n)$ when divided by 3, i.e., $r(n+8)=r(n)$.
Therefore, $r(1996)=r(4+249 \times 8)=r(4)$.
And $r(3)+r(1)+1=4$,
so $r(4)=1$,

which means the remainder when $f(1996)$ is divided by 3 is 1. Therefore, $f(1996)$ is not divisible by 3.