Bulgaria competition problems, 2022

It follows from $d_i \ge i$ and $d_6 d_{2017} = d_4 d_{2019} = n$ that
$$2n = 6 d_{2017} + 4 d_{2019} \le d_6 d_{2017} + d_4 d_{2019} = 2n.$$
Therefore $d_6 = 6$ implying $d_i = i$ for $i = 1, 2, 3, 4, 5, 6$ and thus $n$ is divisible by $2^2 \cdot 3 \cdot 5$. Since $2022 = 2 \cdot 3 \cdot 337$ and $337$ is a prime number it follows that $n = 2^x \cdot 3^y \cdot 5^z$ for $x \ge 2$. All nice numbers are:
$$2^2 \cdot 3 \cdot 5^{336},\quad 2^2 \cdot 3^{336} \cdot 5,\quad 2^{336} \cdot 3 \cdot 5^2,\quad 2^{336} \cdot 3^2 \cdot 5.$$