$6 \cdot 67$ represents any permutation of the integers $1,2, \cdots, n$ using $a_{1}, a_{2}, \cdots, a_{n}$. Let $f(n)$ be the number of permutations that satisfy the conditions:
(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] Clearly, $f(1)=f(2)=1, f(3)=2$.
Let $n \geqslant 4$. Since $a_{1}=1$, then $a_{2}=2$ or 3.
If $a_{2}=2$, the number of permutations of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ is equal to the number of permutations of $a_{2}-1, a_{3}-1, \cdots, a_{n}-1$. The latter is a permutation of $1,2, \cdots, n-1$ and satisfies
$$
a_{2}-1=1
$$

and $\quad\left|\left(a_{i}-1\right)-\left(a_{i+1}-1\right)\right| \leqslant 2, i=2,3, \cdots, n-1$.
Therefore, the number of permutations of the latter is $f(n-1)$, so the number of permutations in the original problem that satisfy $a_{2}=2$ is $f(n-1)$.

If $a_{2}=3, a_{3}=2$, then $a_{4}=4$ must hold. Similarly, the number of permutations in the original problem that satisfy $a_{2}=3$ and $a_{3}=2$ is $f(n-3)$.
If $a_{2}=3, a_{3} \neq 2$, then 2 must be after 4, and no number can follow 2 immediately, so the last two numbers must be 4,2. Thus, 5 must be after 3, and 6 must be before 4. Proceeding in this manner, there is only one such permutation:
$$
1,3,5, \cdots, 6,4,2 \text {. }
$$
(First, arrange the odd numbers in order, then arrange the even numbers in reverse order).
Therefore, we have
$$
f(n)=f(n-1)+f(n-3)+1 \text {. }
$$

Let $f(n)$ modulo 3 be $r(n)$, then
$$
\begin{array}{l}
r(1)=r(2)=1, r(3)=2, \\
r(n)=r(n-1)+r(n-3)+1, n \geqslant 4 .
\end{array}
$$

Thus, the sequence $r(1), r(2), r(3), r(4), \cdots \cdots$ forms a periodic sequence with a period of 8: 1,1, 2,1,0,0,2,0,1,1, \cdots \cdots.
Since $1996=8 \times 249+4$, the remainder when $f(1996)$ is divided by 3 is
$$
r(1996)=r(4)=1 \text {, }
$$

which means $f(1996)$ is not divisible by 3.