imc

A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have: $$\begin{array}{ccccccc}\mathrm{Factorial}& 0!-4!& 5!-9!& 10!-14!& 15!-19!& 20!-24!& 25!-29!\\ \mathrm{Zeros}& 0& 1& 2& 3& 4& 6\end{array}$$ We first look at the case when $n!$ has $1$ zero and $(2n)!$ has $3$ zeros. If $n=5,6,7$, $(2n)!$ has only $2$ zeros. But for $n=8,9$, $(2n)!$ has $3$ zeros. Thus, $n=8$ and $n=9$ work. Secondly, we look at the case when $n!$ has $2$ zeros and $(2n)!$ has $6$ zeros. If $n=10,11,12$, $(2n)!$ has only $4$ zeros. But for $n=13,14$, $(2n)!$ has $6$ zeros. Thus, the smallest four values of $n$ that work are $n=8,9,13,14$, which sum to $44$. The sum of the digits of $44$ is $\boxed{8}$