Find all pairs of integers $(m, n)$ such that $(m - n)^2 =\frac{4mn}{m + n - 1}$
1. Let \( m - n = a \) and \( m + n = b \). Then, we have:
   \[
   (m - n)^2 = a^2 \quad \text{and} \quad m + n = b
   \]
   The given equation is:
   \[
   (m - n)^2 = \frac{4mn}{m + n - 1}
   \]
   Substituting \( a \) and \( b \) into the equation, we get:
   \[
   a^2 = \frac{4mn}{b - 1}
   \]

2. We know that:
   \[
   mn = \frac{b^2 - a^2}{4}
   \]
   Substituting this into the equation, we get:
   \[
   a^2 = \frac{4 \cdot \frac{b^2 - a^2}{4}}{b - 1}
   \]
   Simplifying, we obtain:
   \[
   a^2 = \frac{b^2 - a^2}{b - 1}
   \]

3. Multiplying both sides by \( b - 1 \), we get:
   \[
   a^2(b - 1) = b^2 - a^2
   \]
   Rearranging terms, we have:
   \[
   a^2 b - a^2 = b^2 - a^2
   \]
   Adding \( a^2 \) to both sides, we get:
   \[
   a^2 b = b^2
   \]
   Dividing both sides by \( b \) (assuming \( b \neq 0 \)), we get:
   \[
   a^2 = b
   \]

4. Therefore, we have:
   \[
   (m - n)^2 = m + n
   \]
   Let \( m + n = k^2 \) for some non-negative integer \( k \). Then, \( m - n = \pm k \). This implies:
   \[
   m = n \pm k
   \]

5. We now consider two cases:

   **Case 1:** \( m = n + k \)
   \[
   m + n = k^2 \implies n + k + n = k^2 \implies 2n + k = k^2 \implies n = \frac{k^2 - k}{2}
   \]
   \[
   m = n + k = \frac{k^2 - k}{2} + k = \frac{k^2 + k}{2}
   \]
   Note that \( k^2 - k \) and \( k^2 + k \) are always even, so \( m \) and \( n \) are both integers for integer \( k \).

   **Case 2:** \( m = n - k \)
   \[
   m + n = k^2 \implies n - k + n = k^2 \implies 2n - k = k^2 \implies n = \frac{k^2 + k}{2}
   \]
   \[
   m = n - k = \frac{k^2 + k}{2} - k = \frac{k^2 - k}{2}
   \]
   Again, \( k^2 - k \) and \( k^2 + k \) are always even, so \( m \) and \( n \) are both integers for integer \( k \).

6. Plugging these solutions into the original equation, we find that they indeed work for \( k \neq 1 \).

Conclusion:
\[
\boxed{(m, n) = \left( \frac{k^2 + k}{2}, \frac{k^2 - k}{2} \right), \left( \frac{k^2 - k}{2}, \frac{k^2 + k}{2} \right) \text{ for non-negative integer } k \neq 1}
\]