Problem 1. Consider $A$, the set of natural numbers with exactly 2019 natural divisors, and for each $n \in A$, we denote

$$
S_{n}=\frac{1}{d_{1}+\sqrt{n}}+\frac{1}{d_{2}+\sqrt{n}}+\ldots+\frac{1}{d_{2019}+\sqrt{n}}
$$

where $d_{1}, d_{2}, \ldots, d_{2019}$ are the natural divisors of $n$.

Determine the maximum value of $S_{n}$ as $n$ ranges over the set $A$.
Solution. Since $2019=3 \cdot 673$ and 673 is prime, it follows that any number in $A$ has one of the forms: $p^{2018}$, where $p$ is a prime number, or $p^{2} \cdot q^{672}$, where $p, q$ are distinct primes. Thus, the smallest number in the set $A$ is $3^{2} \cdot 2^{672}$.

$2 p$

Since $d_{i}$ is a divisor of $n \Leftrightarrow \frac{n}{d_{i}}$ is a divisor of $n$, we obtain that

$$
2 \cdot S_{n}=\sum_{i=1}^{2019}\left(\frac{1}{d_{i}+\sqrt{n}}+\frac{1}{\frac{n}{d_{i}}+\sqrt{n}}\right)=\sum_{i=1}^{2019}\left(\frac{1}{d_{i}+\sqrt{n}}+\frac{1}{\sqrt{n}} \cdot \frac{d_{i}}{d_{i}+\sqrt{n}}\right)=\frac{2019}{\sqrt{n}}
$$

It follows that $S_{n}=\frac{2019}{2 \cdot \sqrt{n}} \leq \frac{3 \cdot 673}{2 \cdot \sqrt{3^{2} \cdot 2^{672}}}=\frac{673}{2^{337}}$, the maximum value of $S_{n}$, when $n$ ranges over $A$. Equality is achieved for $n=3^{2} \cdot 2^{672}$.

$1 p$