**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 \( 3^2 \cdot 2672 \) Since \( d_i \) divides \( n \) if and only if \( \frac{n}{d_i} \) divides \( n \), 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 results that \[ S_n = \frac{2019}{2} \cdot \frac{1}{\sqrt{n}} \leq \frac{3 \cdot 673}{2 \cdot \sqrt{3^2 \cdot 2672}} = \frac{673}{2337}, \]the maximum value of \( S_n \) when \( n \) belongs to \( A \). Equality is reached for \( n = 3^2 \cdot 2672 \).