Determine all positive integers $n$ for which there exist positive divisors $a$, $b$, $c$ of $n$ such that $a>b>c$ and $a^2 - b^2$, $b^2 - c^2$, $a^2 - c^2$ are also divisors of $n$.
1. **Verification of \( n = 60k \) for some positive integer \( k \):**

   We start by verifying that \( n = 60k \) works. Consider \( a = 4 \), \( b = 2 \), and \( c = 1 \). We need to check that \( a^2 - b^2 \), \( b^2 - c^2 \), and \( a^2 - c^2 \) are divisors of \( n \).

   \[
   a^2 - b^2 = 4^2 - 2^2 = 16 - 4 = 12
   \]
   \[
   b^2 - c^2 = 2^2 - 1^2 = 4 - 1 = 3
   \]
   \[
   a^2 - c^2 = 4^2 - 1^2 = 16 - 1 = 15
   \]

   Since \( 12 \mid 60k \), \( 3 \mid 60k \), and \( 15 \mid 60k \), it follows that \( n = 60k \) works.

2. **Proving that \( n \) must be of the form \( 60k \):**

   We need to show that any \( n \) satisfying the conditions must be divisible by 3, 4, and 5.

   - **Divisibility by 3:**

     Suppose \( n \) is not divisible by 3. Then \( a^2 \equiv b^2 \equiv 1 \pmod{3} \) because the squares of numbers not divisible by 3 are congruent to 1 modulo 3. This implies:

     \[
     3 \mid a^2 - b^2 \implies 3 \mid n
     \]

     This is a contradiction, so \( n \) must be divisible by 3.

   - **Divisibility by 4:**

     By the pigeonhole principle, at least two of \( a, b, c \) must have the same parity. Without loss of generality, assume \( a \) and \( b \) have the same parity. Then both \( a + b \) and \( a - b \) are even, so:

     \[
     4 \mid (a + b)(a - b) = a^2 - b^2 \implies 4 \mid n
     \]

     Therefore, \( n \) must be divisible by 4.

   - **Divisibility by 5:**

     The squares of integers modulo 5 are 0, 1, or 4. Since \( n \) is not divisible by 5, none of \( a, b, c \) can be divisible by 5. By the pigeonhole principle, at least two of \( a^2, b^2, c^2 \) must be congruent modulo 5. Without loss of generality, assume \( a^2 \equiv b^2 \pmod{5} \). This implies:

     \[
     5 \mid a^2 - b^2 \implies 5 \mid n
     \]

     Therefore, \( n \) must be divisible by 5.

3. **Conclusion:**

   Since \( n \) is divisible by 3, 4, and 5, it follows that:

   \[
   60 = 3 \cdot 4 \cdot 5 \mid n \implies n = 60k \text{ for some positive integer } k
   \]

The final answer is \( \boxed{ n = 60k } \) for some positive integer \( k \).