Determine all positive integers $n \ge 2$ which have a positive divisor $m | n$ satisfying $$n = d^3 + m^3.$$
where $d$ is the smallest divisor of $n$ which is greater than $1$.
1. We start with the given integer \( n \geq 2 \) and a positive divisor \( m \) of \( n \) such that 
   \[
   n = d^3 + m^3,
   \]
   where \( d \) is the smallest divisor of \( n \) greater than 1.

2. Clearly, \( d \) is the smallest prime divisor of \( n \). Hence, \( d = p \), where \( p \) is a prime. Substituting \( d = p \) into the equation, we get
   \[
   n = p^3 + m^3.
   \]

3. Since \( p \mid n \) and \( m \mid n \), it follows that \( p \mid m^3 \) and \( m \mid p^3 \). Therefore, \( m \) must be of the form \( p^k \) for some integer \( k \) such that \( 1 \leq k \leq 3 \).

4. Substituting \( m = p^k \) into the equation, we obtain
   \[
   n = p^3 + p^{3k}.
   \]

5. We now consider the possible values of \( p \) and \( k \):
   - If \( p \) is odd, then \( n \) would be even (since \( p^3 + p^{3k} \) is even for odd \( p \)), which implies \( p = 2 \). This is a contradiction because \( p \) cannot be both odd and equal to 2.
   - Therefore, \( p \) must be even, and the only even prime is \( p = 2 \).

6. Substituting \( p = 2 \) into the equation, we get
   \[
   n = 2^3 + 2^{3k} = 8 + 8^k.
   \]

7. We now evaluate \( n \) for the possible values of \( k \):
   - For \( k = 1 \):
     \[
     n = 8 + 8^1 = 8 + 8 = 16.
     \]
   - For \( k = 2 \):
     \[
     n = 8 + 8^2 = 8 + 64 = 72.
     \]
   - For \( k = 3 \):
     \[
     n = 8 + 8^3 = 8 + 512 = 520.
     \]

8. Therefore, the integers \( n \) that satisfy the given conditions are \( n = 16, 72, \) and \( 520 \).

The final answer is \( \boxed{16, 72, 520} \).