Determine all pairs of positive integers $(m, n)$ for which there exists a bijective function \[f : \mathbb{Z}_m \times \mathbb{Z}_n \to \mathbb{Z}_m \times \mathbb{Z}_n\]such that the vectors $f(\mathbf{v}) + \mathbf{v}$, as $\mathbf{v}$ runs through all of $\mathbb{Z}_m \times \mathbb{Z}_n$, are pairwise distinct. 

(For any integers $a$ and $b$, the vectors $[a, b], [a + m, b]$ and $[a, b + n]$ are treated as equal.)

[i]Poland, Wojciech Nadara[/i]
To determine all pairs of positive integers \((m, n)\) for which there exists a bijective function 
\[f : \mathbb{Z}_m \times \mathbb{Z}_n \to \mathbb{Z}_m \times \mathbb{Z}_n\]
such that the vectors \(f(\mathbf{v}) + \mathbf{v}\), as \(\mathbf{v}\) runs through all of \(\mathbb{Z}_m \times \mathbb{Z}_n\), are pairwise distinct, we need to analyze the properties of such pairs.

1. **Opposite Parity Case:**
   - Assume \(m\) is even and \(n\) is odd.
   - The first entry of the sum of all vectors in \(\mathbb{Z}_m \times \mathbb{Z}_n\) is:
     \[
     (0 + 1 + \cdots + (m-1)) \cdot n = n \cdot \frac{m(m-1)}{2} \equiv \frac{m}{2} \pmod{m}.
     \]
   - If a suitable \(f\) exists, the first entry of the sum of \(f(\mathbf{v}) + \mathbf{v}\) would be:
     \[
     2 \cdot \frac{m}{2} \equiv 0 \pmod{m},
     \]
     which is a contradiction since \(\frac{m}{2}\) is not an integer when \(m\) is even and \(n\) is odd.

2. **Both Entries Odd:**
   - Let \(m\) and \(n\) be both odd.
   - The identity function \(f\) of \(\mathbb{Z}_m \times \mathbb{Z}_n\) is suitable because:
     \[
     f(x, y) + (x, y) = (2x \bmod m, 2y \bmod n), \quad (x, y) \in \mathbb{Z}_m \times \mathbb{Z}_n,
     \]
     are pairwise distinct due to \(m\) and \(n\) being both odd.

3. **Both Entries Even:**
   - Let \(m\) and \(n\) be both even.
   - If \(m = 2\), a suitable bijection \(f : \mathbb{Z}_2 \times \mathbb{Z}_n \to \mathbb{Z}_2 \times \mathbb{Z}_n\) can be defined as:
     \[
     f(0, r) = (0, r), \quad f(1, r) = \left(0, r + \frac{n}{2}\right), \quad f\left(0, r + \frac{n}{2}\right) = (1, r + 1), \quad f\left(1, r + \frac{n}{2}\right) = \left(1, r + \frac{n}{2} + 1\right),
     \]
     for \(0 \leq r < n/2\).

4. **Lemma for Composite Pairs:**
   - If \((a, b)\) and \((c, d)\) are both good, then so is \((ac, bd)\).
   - Given suitable bijections \(f\) and \(g\) for the pairs \((a, b)\) and \((c, d)\), respectively, we can construct a suitable bijection \(h\) for the pair \((ac, bd)\) as:
     \[
     h((ax_1, by_1) + (x_2, y_2)) = (ag_1(x_1, y_1), bg_2(x_1, y_1)) + f(x_2, y_2).
     \]
   - This bijection \(h\) is well-defined and suitable.

5. **General Case for Even Pairs:**
   - Write \(m = 2^k a\) and \(n = 2^\ell b\), where \(k\) and \(\ell\) are positive integers, and \(a\) and \(b\) are odd.
   - Apply the lemma repeatedly for the string of pairs:
     \[
     \underbrace{(2, 2), (2, 2), \ldots, (2, 2)}_{k-1}, (2, 2^{\ell-k+1}), (a, b),
     \]
     each of which falls in one of the previously considered cases.

Conclusion:
The required pairs \((m, n)\) are precisely those having both entries of the same parity.

The final answer is \( \boxed{ (m, n) } \) such that both \(m\) and \(n\) are either both odd or both even.