Think! 2

Every zero at the end of $n!$ corresponds to a factor 5. (It is easy to see that there are always more factors 2 than 5 in $n!$ )

In the product $n! = 1\cdot 2\cdot 3 \cdots n$ , every fifth number contributes a factor 5; every twenty-fifth number contributes an additional factor 5; and so on. Therefore the number of factors 5 is $\left\lfloor \frac n 5\right\rfloor + \left\lfloor \frac n {25}\right\rfloor + \left\lfloor \frac n {125}\right\rfloor + \cdots$ Ignoring the $\lfloor\cdot\rfloor$ , we find that the number of factors 5 is approximately (and no greater than) $n\cdot \left(\frac15 + \frac1{25} + \frac1{125} + \cdots\right) = \frac n 4.$ If there are 2020 zeroes, and therefore 2020 factors 5, we expect $n \approx 4\cdot 2020 = 8080.$ In fact, for 8080 we find that there are $\left\lfloor \frac {8080}5\right\rfloor + \left\lfloor \frac {8080}{25}\right\rfloor + \left\lfloor \frac {8080}{125}\right\rfloor + \left\lfloor \frac {8080}{625}\right\rfloor + \left\lfloor \frac {8080}{3125}\right\rfloor = 1616 + 323 + 64 + 12 + 2 = 2017$ factors 5. The remaining three factors 5 come from 8085, 8090, and 8095.

Since the next factor 5 is found at 8100, the required values of $n$ are $n = 8095, 8096, 8097, 8098, 8099.$ The sum is easily found as five times the middle value, $\sum n = 5\cdot 8097 = \boxed{40485}.$