4. Does there exist a natural number $n$, such that the first four digits of $n!$ are 1993?
4. Existence. Let $m=1000100000$. When $k<99999$, if $(m+k)!=\overline{a b c d} \cdots$, then $(m+k+1)!=(m+k)!\times (m+k+1)=\overline{a b c d} \times 10001 \cdots=\overline{a b c x \cdots}$, where $x=d$ or $d+1$. Therefore, if $n!=\overline{a b c d \cdots}$, then the first four digits of $(m+1)!,(m+2)!, \cdots,(m+99999)!$ are the same as or one more than the previous one. Moreover (due to the increase in the fifth digit from the left), the first four digits must increase by 1 after at most 10 numbers. Thus, the first four digits of the 100000 numbers $m!,(m+1)!, \cdots,(m+99999)!$ cover all 10000 values, and among them, 1993 must appear.