Determine all positive integers $n$ such that $ xy+1 \equiv 0 \; \pmod{n} $ implies that $ x+y \equiv 0 \; \pmod{n}$.
To determine all positive integers \( n \) such that \( xy + 1 \equiv 0 \pmod{n} \) implies \( x + y \equiv 0 \pmod{n} \), we proceed as follows:

1. **Assume \( xy + 1 \equiv 0 \pmod{n} \)**:
   \[
   xy \equiv -1 \pmod{n}
   \]

2. **Consider \( x \) and \( y \) such that \( x \) is coprime to \( n \)**:
   Since \( x \) is coprime to \( n \), there exists an integer \( x^{-1} \) such that:
   \[
   x x^{-1} \equiv 1 \pmod{n}
   \]

3. **Express \( y \) in terms of \( x \)**:
   From \( xy \equiv -1 \pmod{n} \), multiply both sides by \( x^{-1} \):
   \[
   y \equiv -x^{-1} \pmod{n}
   \]

4. **Check the condition \( x + y \equiv 0 \pmod{n} \)**:
   Substitute \( y \) from the previous step:
   \[
   x + y \equiv x - x^{-1} \equiv 0 \pmod{n}
   \]
   This implies:
   \[
   x \equiv x^{-1} \pmod{n}
   \]

5. **Analyze the implications of \( x \equiv x^{-1} \pmod{n} \)**:
   This means \( x^2 \equiv 1 \pmod{n} \). Therefore, \( x \) must be a solution to the equation:
   \[
   x^2 - 1 \equiv 0 \pmod{n}
   \]
   \[
   (x - 1)(x + 1) \equiv 0 \pmod{n}
   \]

6. **Determine the structure of \( n \)**:
   For the above equation to hold for all \( x \) coprime to \( n \), \( n \) must be such that every \( x \) coprime to \( n \) satisfies \( x^2 \equiv 1 \pmod{n} \). This is true if and only if \( n \) is a product of distinct primes, specifically \( n = 2 \).

7. **Verify \( n = 2 \)**:
   If \( n = 2 \), then:
   \[
   xy + 1 \equiv 0 \pmod{2} \implies xy \equiv -1 \equiv 1 \pmod{2}
   \]
   Since \( x \) and \( y \) are integers, \( x \) and \( y \) must both be odd, and thus:
   \[
   x + y \equiv 1 + 1 \equiv 0 \pmod{2}
   \]
   This satisfies the condition \( x + y \equiv 0 \pmod{2} \).

Therefore, the only positive integer \( n \) that satisfies the given condition is \( n = 2 \).

The final answer is \( \boxed{2} \).