Determine all positive integers $n>2$, such that $n = a^3 + b^3$, where $a$ is the smallest positive divisor of $n$ greater than $1$ and $b$ is an arbitrary positive divisor of $n$. 
1. Given \( n = a^3 + b^3 \), where \( a \) is the smallest positive divisor of \( n \) greater than 1 and \( b \) is an arbitrary positive divisor of \( n \).

2. Since \( a \) is the smallest positive divisor of \( n \) greater than 1, \( a \) must be a prime number. This is because the smallest divisor greater than 1 of any number is a prime number.

3. We need to consider the possible values of \( b \). Since \( b \) is a divisor of \( n \), \( b \) can be \( a \), \( a^2 \), or \( a^3 \).

4. If \( a > 2 \), then \( a \) is an odd prime number. This implies that \( a^3 \) is also odd. Since \( b \) is a divisor of \( n \), \( b \) must also be odd. Therefore, \( b^3 \) is odd. The sum of two odd numbers is even, so \( a^3 + b^3 \) is even. However, if \( a > 2 \), then \( n \) is odd, which leads to a contradiction because \( n \) must be even. Therefore, \( a \) cannot be greater than 2.

5. Hence, \( a = 2 \). Now we need to find the possible values of \( n \) when \( a = 2 \).

6. If \( a = 2 \), then \( n = 2^3 + b^3 \). The possible values of \( b \) are \( 2 \), \( 2^2 \), and \( 2^3 \).

7. For \( b = 2 \):
   \[
   n = 2^3 + 2^3 = 8 + 8 = 16
   \]

8. For \( b = 2^2 \):
   \[
   n = 2^3 + (2^2)^3 = 8 + 64 = 72
   \]

9. For \( b = 2^3 \):
   \[
   n = 2^3 + (2^3)^3 = 8 + 512 = 520
   \]

10. Therefore, the possible values of \( n \) are \( 16 \), \( 72 \), and \( 520 \).

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