Determine all positive integers $n \ge 2$ that satisfy the following condition; For all integers $a, b$ relatively prime to $n$, \[a \equiv b \; \pmod{n}\Longleftrightarrow ab \equiv 1 \; \pmod{n}.\]
1. **Understanding the given condition:**
   We need to find all positive integers \( n \ge 2 \) such that for all integers \( a \) and \( b \) that are relatively prime to \( n \), the following equivalence holds:
   \[
   a \equiv b \pmod{n} \Longleftrightarrow ab \equiv 1 \pmod{n}.
   \]
   This means that if \( a \equiv b \pmod{n} \), then \( ab \equiv 1 \pmod{n} \), and vice versa.

2. **Analyzing the condition:**
   Let \( a \equiv a \pmod{n} \). According to the given condition, this implies:
   \[
   a^2 \equiv 1 \pmod{n}.
   \]
   This must hold for all \( a \) that are coprime to \( n \).

3. **Considering specific values:**
   Let's consider \( a = 2 \). For \( n \) to satisfy the condition, \( 2^2 \equiv 1 \pmod{n} \) must hold. This implies:
   \[
   4 \equiv 1 \pmod{n} \implies 3 \equiv 0 \pmod{n} \implies n \mid 3.
   \]
   Therefore, \( n \) must be a divisor of 3. The divisors of 3 are 1 and 3. Since \( n \ge 2 \), we have \( n = 3 \).

4. **Checking \( n = 3 \):**
   For \( n = 3 \), we need to verify if the condition holds. Let \( a \) and \( b \) be integers coprime to 3. The integers coprime to 3 are 1 and 2.
   - If \( a \equiv b \pmod{3} \), then \( a = b \).
   - We need to check if \( ab \equiv 1 \pmod{3} \):
     - For \( a = 1 \), \( b = 1 \), \( 1 \cdot 1 \equiv 1 \pmod{3} \).
     - For \( a = 2 \), \( b = 2 \), \( 2 \cdot 2 \equiv 4 \equiv 1 \pmod{3} \).

   Thus, \( n = 3 \) satisfies the condition.

5. **Considering other values of \( n \):**
   We need to check if there are other values of \( n \) that satisfy the condition. Let's consider \( n = 8 \):
   - For \( n = 8 \), the integers coprime to 8 are 1, 3, 5, and 7.
   - We need to check if \( a^2 \equiv 1 \pmod{8} \) for all \( a \) coprime to 8:
     - For \( a = 1 \), \( 1^2 \equiv 1 \pmod{8} \).
     - For \( a = 3 \), \( 3^2 \equiv 9 \equiv 1 \pmod{8} \).
     - For \( a = 5 \), \( 5^2 \equiv 25 \equiv 1 \pmod{8} \).
     - For \( a = 7 \), \( 7^2 \equiv 49 \equiv 1 \pmod{8} \).

   Thus, \( n = 8 \) also satisfies the condition.

6. **Considering the Chinese Remainder Theorem:**
   By the Chinese Remainder Theorem, if \( n = 3 \) and \( n = 8 \) both satisfy the condition, then \( n = 24 \) (the least common multiple of 3 and 8) also satisfies the condition.

7. **Checking divisors of 24:**
   The divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. We have already checked \( n = 3 \) and \( n = 8 \). We need to check the other divisors:
   - For \( n = 4 \), the integers coprime to 4 are 1 and 3. We need to check if \( a^2 \equiv 1 \pmod{4} \):
     - For \( a = 1 \), \( 1^2 \equiv 1 \pmod{4} \).
     - For \( a = 3 \), \( 3^2 \equiv 9 \equiv 1 \pmod{4} \).

   Thus, \( n = 4 \) satisfies the condition.

   - For \( n = 6 \), the integers coprime to 6 are 1 and 5. We need to check if \( a^2 \equiv 1 \pmod{6} \):
     - For \( a = 1 \), \( 1^2 \equiv 1 \pmod{6} \).
     - For \( a = 5 \), \( 5^2 \equiv 25 \equiv 1 \pmod{6} \).

   Thus, \( n = 6 \) satisfies the condition.

   - For \( n = 12 \), the integers coprime to 12 are 1, 5, 7, and 11. We need to check if \( a^2 \equiv 1 \pmod{12} \):
     - For \( a = 1 \), \( 1^2 \equiv 1 \pmod{12} \).
     - For \( a = 5 \), \( 5^2 \equiv 25 \equiv 1 \pmod{12} \).
     - For \( a = 7 \), \( 7^2 \equiv 49 \equiv 1 \pmod{12} \).
     - For \( a = 11 \), \( 11^2 \equiv 121 \equiv 1 \pmod{12} \).

   Thus, \( n = 12 \) satisfies the condition.

   - For \( n = 24 \), the integers coprime to 24 are 1, 5, 7, 11, 13, 17, 19, and 23. We need to check if \( a^2 \equiv 1 \pmod{24} \):
     - For \( a = 1 \), \( 1^2 \equiv 1 \pmod{24} \).
     - For \( a = 5 \), \( 5^2 \equiv 25 \equiv 1 \pmod{24} \).
     - For \( a = 7 \), \( 7^2 \equiv 49 \equiv 1 \pmod{24} \).
     - For \( a = 11 \), \( 11^2 \equiv 121 \equiv 1 \pmod{24} \).
     - For \( a = 13 \), \( 13^2 \equiv 169 \equiv 1 \pmod{24} \).
     - For \( a = 17 \), \( 17^2 \equiv 289 \equiv 1 \pmod{24} \).
     - For \( a = 19 \), \( 19^2 \equiv 361 \equiv 1 \pmod{24} \).
     - For \( a = 23 \), \( 23^2 \equiv 529 \equiv 1 \pmod{24} \).

   Thus, \( n = 24 \) satisfies the condition.

Conclusion:
The positive integers \( n \ge 2 \) that satisfy the given condition are \( n = 3, 4, 6, 8, 12, 24 \).

The final answer is \( \boxed{ n = 3, 4, 6, 8, 12, 24 } \).