For a positive integer $n$, sort its positive divisors as $1=d_{1}<d_{2}<\cdots<d_{k}=n$. Find all positive integers $n$ that are divisible by 2019 and satisfy $n=d_{19} d_{20}$.
5. Notice that, $n=d_{19} d_{20}$.

Thus, $n$ has exactly 38 positive divisors, and
$$
n=d_{19} d_{20}=d_{18} d_{21}=d_{17} d_{22}=\cdots=d_{1} d_{38}=d_{38} \text {. }
$$

Then $n=p^{\alpha}$, where $p$ is a prime and $\alpha+1=38$, at this time, $n=p^{37}$ or $n=p^{\alpha} q^{\beta}$, where $p, q$ are two different primes and
$$
\begin{array}{l}
(\alpha+1)(\beta+1)=38 \\
\Rightarrow(\alpha, \beta)=(1,18) \text { or }(18,1),
\end{array}
$$

at this time, $n=p q^{18}$ or $n=p^{18} q$.
Now consider 2019|n.
Since $2019=3 \times 673$, and $3, 673$ are both primes, so, $n=3 \times 673^{18}$ or $n=3^{18} \times 673$.
Next, verify whether it satisfies $n=d_{19} d_{20}$.
(1) If $n=3 \times 673^{18}$, arranging all positive divisors in order gives
$$
\begin{array}{l}
1<3<673<3 \times 673<673^{2}<3 \times 673^{2} \\
<\cdots<673^{9}<3 \times 673^{9}<\cdots
\end{array}
$$

It is easy to see that, $d_{19}=673^{9}, d_{20}=3 \times 673^{9}$, satisfying
$$
n=d_{19} d_{20} \text {. }
$$
(2) If $n=3^{18} \times 673$, arranging all positive divisors in order gives
$$
\begin{array}{l}
1<3<3^{2}<3^{3}<3^{4}<3^{5}<673<3^{6} \\
<3 \times 673<3^{7}<3^{2} \times 673<3^{8} \\
<3^{3} \times 673<3^{9}<3^{4} \times 673<3^{10} \\
<3^{5} \times 673<3^{11}<3^{6} \times 673<3^{12}<\cdots
\end{array}
$$

It is easy to see that, $d_{19}=3^{6} \times 673, d_{20}=3^{12}$, satisfying $n=d_{19} d_{20}$.
Therefore, $n=3 \times 673^{18}$ or $n=3^{18} \times 673$.