Let $n = p_1^{a_1} \cdots p_k^{a_k}$ be the prime factorization of $n$. Let $d = p_1^{b_1} \cdots p_k^{b_k}$ be the prime factorization of a divisor $d$ of $n$. Therefore, $0 \le b_i \le a_i$ for all $1\le i \le k$. Further $d$ is a perfect square if and only if $b_i$ is even for all $1\le i \le k$, and $d$ is a perfect cube if and only if $b_i$ is a multiple of $3$ for all $1\le i \le k$. Thus, $$D_2(n) = \prod_{i =1}^k \left( \left \lfloor \frac{a_i}{2} \right \rfloor + 1 \right), D_3(n) = \prod_{i =1}^k \left( \left \lfloor \frac{a_i}{3} \right \rfloor + 1 \right) $$Let $d_2(a) = \left \lfloor \frac{a}{2} \right \rfloor + 1$ and $d_3(a) = \left \lfloor \frac{a}{3} \right \rfloor + 1$, then 1. $d_2(2) = 2, d_3(2) = 1$. Therefore we can multiply by a factor of $2$. 2. $d_2(4) = 3, d_3(4) = 2$. Therefore we can multiply by a factor of $\frac 32$. 3. $d_2(8) = 5, d_3(8) = 3$. Therefore we can multiply by a factor of $\frac 53$. 3. $d_2(72) = 37, d_3(72) = 25$. Therefore we can multiply by a factor of $\frac{37}{25}$. Now, $999 = 37 \cdot 3 \cdot 3 \cdot 3 = \frac{37}{25} \cdot \left(\frac 53 \right)^2 \cdot \left(\frac 32 \right)^5 \cdot 2^5.$ Thus, $n = p_1^{37} \cdot \prod_{i=2}^3 p_i^8 \cdot \prod_{i=4}^8 p_i^4 \cdot \prod_{i=9}^{13} p_i^2$ satisfies that $D_2(n) = 999 \cdot D_3(n)$, where $p_1, \ldots, p_{13}$ are any $13$ distinct primes.