Given $n \ge 2$ distinct positive integers $a_1, a_2, ..., a_n$ none of which is a perfect cube. Find the maximal possible number of perfect cubes among their pairwise products.
To solve this problem, we need to determine the maximal number of perfect cubes among the pairwise products of \( n \) distinct positive integers \( a_1, a_2, \ldots, a_n \), none of which is a perfect cube. We will use properties of exponents and modular arithmetic to analyze the conditions under which the product of two numbers is a perfect cube.

1. **Prime Factorization and Exponent Analysis:**
   For any integer \( a_i \), let \( v_p(a_i) \) denote the exponent of the prime \( p \) in the prime factorization of \( a_i \). We can write:
   \[
   a_i = p_1^{e_{i1}} p_2^{e_{i2}} \cdots p_k^{e_{ik}}
   \]
   where \( p_1, p_2, \ldots, p_k \) are distinct primes and \( e_{ij} \) are the corresponding exponents.

2. **Condition for Perfect Cube:**
   A number \( a_i a_j \) is a perfect cube if and only if for every prime \( p \) dividing \( a_i a_j \), the sum of the exponents of \( p \) in \( a_i \) and \( a_j \) is divisible by 3. That is:
   \[
   v_p(a_i) + v_p(a_j) \equiv 0 \pmod{3}
   \]

3. **Case 1: Existence of \( t(i)_p \equiv 0 \pmod{3} \):**
   Suppose there exists an \( i \) and a prime \( p \) such that \( v_p(a_i) \equiv 0 \pmod{3} \). If \( a_i \) is paired with any \( a_j \) such that \( v_p(a_j) \equiv 0 \pmod{3} \), then \( a_i a_j \) will be a perfect cube with respect to \( p \). However, since none of the \( a_i \) are perfect cubes, we can assume that no \( v_p(a_i) \equiv 0 \pmod{3} \).

4. **Case 2: No \( t(i)_p \equiv 0 \pmod{3} \):**
   Assume that for all \( i \) and all primes \( p \) dividing \( a_i \), \( v_p(a_i) \not\equiv 0 \pmod{3} \). This means \( v_p(a_i) \) can only be \( 1 \) or \( 2 \) modulo 3.

5. **Partitioning into Sets:**
   For each prime \( p \), define two sets:
   \[
   A_1 = \{ i \mid v_p(a_i) \equiv 1 \pmod{3} \}
   \]
   \[
   A_2 = \{ i \mid v_p(a_i) \equiv 2 \pmod{3} \}
   \]
   The product \( a_i a_j \) will be a perfect cube if \( i \in A_1 \) and \( j \in A_2 \) or vice versa.

6. **Counting Perfect Cubes:**
   The number of perfect cubes among the pairwise products is maximized when the sizes of \( A_1 \) and \( A_2 \) are as balanced as possible. Let \( |A_1| = k \) and \( |A_2| = n - k \). The number of perfect cubes is:
   \[
   |A_1||A_2| = k(n - k)
   \]
   This expression is maximized when \( k = \frac{n}{2} \). Therefore, the maximum number of perfect cubes is:
   \[
   \left( \frac{n}{2} \right) \left( \frac{n}{2} \right) = \frac{n^2}{4}
   \]

Conclusion:
The maximal possible number of perfect cubes among the pairwise products of \( n \) distinct positive integers, none of which is a perfect cube, is \( \frac{n^2}{4} \).

The final answer is \(\boxed{\frac{n^2}{4}}\).