Suppose $r \ge 2$ is an integer, and let $m_1, n_1, m_2, n_2, \dots, m_r, n_r$ be $2r$ integers such that $$\left|m_in_j-m_jn_i\right|=1$$ for any two integers $i$ and $j$ satisfying $1 \le i<j \le r$. Determine the maximum possible value of $r$. 

[i]Proposed by B Sury[/i]
1. We start by noting the given condition: for any two integers \(i\) and \(j\) such that \(1 \le i < j \le r\), we have \(\left|m_i n_j - m_j n_i\right| = 1\). This implies that the determinant of the matrix formed by \((m_i, n_i)\) and \((m_j, n_j)\) is \(\pm 1\).

2. We need to determine the maximum possible value of \(r\). Let's analyze the parity of \(m_i\) and \(n_i\).

3. **Case 1**: Both \(m_1\) and \(n_1\) are odd.
   - If \(m_1\) and \(n_1\) are both odd, then for \(\left|m_1 n_j - m_j n_1\right| = 1\) to hold, \(m_j\) and \(n_j\) cannot both be odd or both be even. Therefore, one of \(m_j\) or \(n_j\) must be even.
   - Suppose \(m_2\) is even and \(n_2\) is odd. Then, for \(i \ge 3\), \(m_i\) must be odd and \(n_i\) must be even to satisfy \(\left|m_i n_2 - m_2 n_i\right| = 1\).
   - Now, consider \(r \ge 4\). We have \(m_3, m_4\) odd and \(n_3, n_4\) even. Then \(\left|m_3 n_4 - m_4 n_3\right| = 1\) must hold. However, since both \(m_3 n_4\) and \(m_4 n_3\) are even, their difference cannot be 1, leading to a contradiction. Hence, \(r \le 3\) in this case.

4. **Case 2**: \(m_1\) and \(n_1\) have different parity.
   - Without loss of generality, assume \(m_1\) is even and \(n_1\) is odd. Then, for \(\left|m_1 n_j - m_j n_1\right| = 1\) to hold, \(m_j\) must be odd for \(j \ge 2\).
   - If \(n_2\) is even, then \(n_j\) must be odd for \(j \ge 2\). For \(r \ge 4\), we have \(m_3, m_4\) odd and \(n_3, n_4\) odd. Then \(\left|m_3 n_4 - m_4 n_3\right| = 1\) must hold. However, since both \(m_3 n_4\) and \(m_4 n_3\) are odd, their difference cannot be 1, leading to a contradiction. Hence, \(r \le 3\) in this case.

5. **Conclusion**: In both cases, we find that \(r \le 3\). Therefore, the maximum possible value of \(r\) is 3.

\(\blacksquare\)

The final answer is \( \boxed{ 3 } \)