Belarusian Mathematical Olympiad

Answer: $n = 3 \cdot 673^{18}$; $n = 3^{18} \cdot 673$.

Note that the equality $n = d_{19} \cdot d_{20}$ implies that $n$ has exactly $38$ divisors. Indeed,
$$n = d_{19} \cdot d_{20} = d_{18} \cdot d_{21} = d_{17} \cdot d_{22} = \dots = d_1 \cdot d_{38} = d_{38}.$$
Hence either $n = p^\alpha$, where $p$ is prime and then $\alpha + 1 = 38$, so $n = p^{37}$, or $n = p^\alpha \cdot q^\beta$ ($p$ and $q$ are different primes) and then $38 = (\alpha + 1)(\beta + 1)$, i.e. $(\alpha; \beta) = (1; 18)$ or $(\alpha; \beta) = (18; 1)$, so $n = p \cdot q^{18}$ or $n = p^{18} \cdot q$, respectively.

Recall that $n$ is divisible by $2019$ while $2019 = 3 \cdot 673$, where $3$ and $673$ are primes. Therefore, $n = 3 \cdot 673^{18}$ or $n = 673 \cdot 3^{18}$. It remains to verify that in both of these cases $n = d_{19} \cdot d_{20}$.

1. Let $n = 3 \cdot 673^{18}$. Arrange the divisors in an ascending order:
$$1 < 3 < 673 < 673 \cdot 3 < 673^2 < 673^2 \cdot 3 < \dots < 673^9 < 673^9 \cdot 3 < \dots$$
It is easy to see that $d_{19} = 673^9$, $d_{20} = 673^9 \cdot 3$ and $n = d_{19} \cdot d_{20}$.

2. Let $n = 3^{18} \cdot 673$, then
$$1 < 3 < 3^2 < 3^3 < 3^4 < 3^5 < 673 < 3^6 < 3 \cdot 673 < 3^7 < 3^2 \cdot 673 < 3^8 < 3^3 \cdot 673 < \\ < 3^9 < 3^4 \cdot 673 < 3^{10} < 3^5 \cdot 673 < 3^{11} < 3^6 \cdot 673 < 3^{12} < \dots$$
It is easy to see that $d_{19} = 3^6 \cdot 673$, $d_{20} = 3^{12}$ and $n = d_{19} \cdot d_{20}$.