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 $\boxed{2}$.