Determine all positive integers $n \geq 2$ for which there exists a positive divisor $m \mid n$ such that

$$
n=d^{3}+m^{3},
$$

where $d$ is the smallest divisor of $n$ greater than 1.
The smallest divisor of $n$ greater than 1 is the smallest prime that is a divisor of $n$, so $d$ is prime. Furthermore, $d \mid n$, so $d \mid d^{3}+m^{3}$, thus $d \mid m^{3}$. This implies that $m>1$. On the other hand, $m \mid n$, so $m \mid d^{3}+m^{3}$, thus $m \mid d^{3}$. Since $d$ is prime and $m>1$, we see that $m$ is equal to $d, d^{2}$, or $d^{3}$.

In all cases, the parity of $m^{3}$ is the same as that of $d^{3}$, so $n=d^{3}+m^{3}$ is even. This means that the smallest divisor of $n$ greater than 1 is equal to 2, so $d=2$. We now find the possible solutions: in the case $m=d$, the number $n=2^{3}+2^{3}=16$; in the case $m=d^{2}$, the number $n=2^{3}+2^{6}=72$; and in the case $m=d^{3}$, the number $n=2^{3}+2^{9}=520$. These indeed satisfy the conditions: they are even, so $d=2$, and $2|16 ; 4| 72$ and $8 \mid 520$, so $m \mid n$.