Solution: Using an obvious fact, that is, "the number of trailing zeros in $m$! = the power of 10 in $m$! = the power of 5 in $m$!".

Let $f(m)$ be the number of trailing zeros in $m$!, then $f(m)$ is a non-decreasing function of $m$, and when $m$ is a multiple of 5, we have $f(m)=f(m+1)=f(m+2)=f(m+3)$ $=f(m+4)1991 \times 4=7964$.
Using formula (2) for $f(m)$, we find $f(7965)=1988, f(7975)=1991$, $f(7979)=1991$, according to table (1) we have
$$
0,0,0,0,0,1,1,1,1,1, \cdots, 1989,1989,1989,1989,1989 \text {, }
$$
$1991,1991,1991,1991,1991$.

The above sequence has a total of 7980 terms, with each number appearing exactly 5 times, so there are $7980 \div 5=$ 1596 numbers taken from the set $\{0,1,2, \cdots, 1991\}$, this set contains 1992 elements, of which the number of elements not taken is $1992-1596=396$.
Therefore, there are 396 positive integers less than 1992 that are not the tail of a factorial.
Example 11 Define a positive integer $n$ to be a "factorial tail" 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 "factorial tails"?