If there exists a positive integer $m$ such that $m!$ ends with exactly $n$ zeros, then the positive integer $n$ is called a "factorial tail". How many numbers less than 1992 are not "factorial tails"?
Let $f(m)$ be the number of trailing zeros in the number $m!$. Clearly, $f(m)$ is a non-decreasing function of $m$, and when $m$ is a multiple of 5, we have
$$
f(m)=f(m+1)=f(m+2)=f(m+3)=f(m+4)4 \times 1991=7964$. Using (2), it is easy to see that $f(7965)=1988, f(7975)=1991$. The sequence (1) from the first term to $f(7979)=1991$, consists of 7980 numbers, i.e.,
$$
\begin{array}{c}
0,0,0,0,0,1,1,1,1,1, \cdots, 1991, \\
1991,1991,1991,1991 .
\end{array}
$$

There are $\frac{7980}{5}=1596$ different numbers. Therefore, there are $1992-1596=396$ positive integers less than 1992 that are not "factorial tail numbers".