$8 \cdot 78$ consists of integers that are divisible by 3 and are 1 less than a perfect square, forming an increasing sequence $3,15,24,48, \cdots$ What is the remainder when the 1994th term of this sequence is divided by 1000?
[Solution] A number that is 1 less than a perfect square has the form $n^{2}-1=(n-1)$ $(n+1), n=2,3, \cdots$

If and only if $n$ is not divisible by 3, $n^{2}-1$ is a multiple of 3. Therefore, the $(2 k-1)$-th and $2 k$-th terms of this sequence are
$$
(3 k-1)^{2}-1 \text { and }(3 k+1)^{2}-1 \text {. }
$$

Thus, the 1994th term of this sequence is
$$
\begin{aligned}
(3 \cdot 997+1)^{2}-1 & =(3000-8)^{2}-1 \\
& =3000^{2}-16 \cdot 3000+63,
\end{aligned}
$$

The remainder when it is divided by 1000 is 63.