jmc

Every 20-pretty integer can be written in form $n=2^a{5}^{b}k$, where $a\ge 2$, $b\ge 1$, $\gcd (k,10)=1$, and $d(n)=20$, where $d(n)$ is the number of divisors of $n$. Thus, we have $20=(a+1)(b+1)d(k)$, using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check. If $a+1=4$, then $b+1=5$. But this leads to no solutions, as $(a,b)=(3,4)$ gives $2^3{5}^4>2019$. If $a+1=5$, then $b+1=2$ or $4$. The first case gives $n=2^4\cdot 5^1\cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p<2019⟹p=3,7,11,13,17,19,23$. The sum of all such $n$ is $80(3+7+11+13+17+19+23)=7440$. In the second case $b+1=4$ and $d(k)=1$, and there is one solution $n=2^4\cdot 5^3=2000$. If $a+1=10$, then $b+1=2$, but this gives $2^9\cdot 5^1>2019$. No other values for $a+1$ work. Then we have $\frac{S}{20}=\frac{80(3+7+11+13+17+19+23)+2000}{20}=372+100=\boxed{472}$.