Consider $A$ the set of natural numbers with exactly 2019 divisors, and for each $n \in A$ denote $$ S_n = \frac{1}{d_1+\sqrt{n}} + \frac{1}{d_2+\sqrt{n}} + ... + \frac{1}{d_{2019}+\sqrt{n}}, $$ where $d_1, d_2, \dots, d_{2019}$ are the divisors of $n$. Find the maximum value of $S_n$.
I started off with the theorem where the number of divisors of a number can be found by adding one to all the exponents and then multiplying them which means each divisor is either $p^{2018}$, $p^2$, or $p^{672}$ because $2019 = 3\times 673$. Thank you for helping out!
Edit: I've made some progress. To minimize the denominators, we must make n as small as possible, and by the # of divisors formula, the smallest possibility would be $$n = 2^{672} * 3^2$$
Now, I just need to figure out how to sum this with that giant equation.