Example 4 Define a positive integer $n$ to be a "tail" of a factorial if there exists a positive integer $m$ such that the decimal representation of $m$! ends with exactly $n$ zeros. How many positive integers less than 1992 are not tails of a factorial?
Solution: Clearly, the number of trailing zeros in $m!$ = the number of times 10 is a factor in $m!$ = the number of times 5 is a factor in $m!$.
Thus, we have $f(m)=\sum_{k=1}^{+\infty}\left[\frac{m}{5^{k}}\right]$.
From 1 CुS $1=\sum_{k=1}^{\infty}\left[\frac{m}{5^{k}}\right]7964$.
It is also easy to see that when $5 \mid m$, $f(m)=f(m+1)$ $=f(m+2)=f(m+3)=f(m+4)<f(m+5)$. Therefore, from (1) we calculate
$$
\begin{array}{l}
f(7965)=1988, \\
f(7970)=1989, \\
f(7975)=1991, \\
f(7979)=1991 .
\end{array}
$$

Thus, for $m=0,1,2, \cdots, 7979$, the values of $f(m)$ are sequentially
$$
0,0,0,0,0,1,1,1,1,1, \cdots, 1989,1989 \text {, }
$$
1989, 1989, 1989, 1991, 1991, 1991, 1991, 1991

There are 7980 terms in total, with each number appearing 5 times, so there are $7980 \div 5=1596$ distinct numbers. Therefore, among the positive integers less than 1992, the number of integers that are not "factorial tails" is $1992-1596=396$.