jmc

Let the number of zeros at the end of $m!$ be $f(m)$. We have $f(m)=\lfloor \frac{m}{5}\rfloor +\lfloor \frac{m}{25}\rfloor +\lfloor \frac{m}{125}\rfloor +\lfloor \frac{m}{625}\rfloor +\lfloor \frac{m}{3125}\rfloor +\cdots$. Note that if $m$ is a multiple of $5$, $f(m)=f(m+1)=f(m+2)=f(m+3)=f(m+4)$. Since $f(m)\le \frac{m}{5}+\frac{m}{25}+\frac{m}{125}+\cdots =\frac{m}{4}$, a value of $m$ such that $f(m)=1991$ is greater than $7964$. Testing values greater than this yields $f(7975)=1991$. There are $\frac{7975}{5}=1595$ distinct positive integers, $f(m)$, less than $1992$. Thus, there are $1991-1595=\boxed{396}$ positive integers less than $1992$ that are not factorial tails.