Find all pairs of positive integers $ m$ and $ n$ that satisfy (both) following conditions:

(i) $ m^{2}\minus{}n$ divides $ m\plus{}n^{2}$ 

(ii) $ n^{2}\minus{}m$ divides $ n\plus{}m^{2}$
To find all pairs of positive integers \( m \) and \( n \) that satisfy the given conditions, we will analyze the problem step-by-step.

1. **Assume \( m \geq n \) and \( m \geq 4 \)**:
   - Given \( m \geq n \) and \( m \geq 4 \), we have \( m^2 \geq n^2 \) and \( m(m - 1) > 2n \) because \( m \geq n \) and \( m - 1 \geq 3 \).
   - Therefore, \( m^2 + m(m - 1) > n^2 + 2n \).

2. **Analyze the first condition \( m^2 - n \mid m + n^2 \)**:
   - Since \( m^2 - n \mid m + n^2 \), we can write:
     \[
     m + n^2 = k(m^2 - n) \quad \text{for some integer } k.
     \]
   - Rearranging, we get:
     \[
     m + n^2 = km^2 - kn.
     \]
   - This implies:
     \[
     m + n^2 + kn = km^2.
     \]
   - Simplifying further:
     \[
     km^2 - kn - n^2 = m.
     \]

3. **Analyze the second condition \( n^2 - m \mid n + m^2 \)**:
   - Since \( n^2 - m \mid n + m^2 \), we can write:
     \[
     n + m^2 = l(n^2 - m) \quad \text{for some integer } l.
     \]
   - Rearranging, we get:
     \[
     n + m^2 = ln^2 - lm.
     \]
   - This implies:
     \[
     n + m^2 + lm = ln^2.
     \]
   - Simplifying further:
     \[
     ln^2 - lm - m^2 = n.
     \]

4. **Assume \( m = n + 1 \)**:
   - From the first condition, substituting \( m = n + 1 \):
     \[
     (n + 1)^2 - n = n^2 + 2n + 1 - n = n^2 + n + 1.
     \]
   - We need \( n^2 + n + 1 \mid (n + 1) + n^2 \):
     \[
     n^2 + n + 1 \mid n^2 + n + 1.
     \]
   - This is always true.

5. **Substitute \( m = n + 1 \) into the second condition**:
   - We need \( n^2 - (n + 1) \mid n + (n + 1)^2 \):
     \[
     n^2 - n - 1 \mid n + n^2 + 2n + 1 = n^2 + 3n + 1.
     \]
   - Simplifying:
     \[
     n^2 - n - 1 \mid 4n + 2.
     \]
   - Since \( n^2 - n - 1 \) is odd, we have:
     \[
     n^2 - n - 1 \mid 2n + 1.
     \]
   - This implies:
     \[
     n^2 - n - 1 \leq 2n + 1 \implies n^2 - 3n - 2 \leq 0.
     \]
   - Solving the quadratic inequality:
     \[
     n^2 - 3n - 2 = 0 \implies (n - 3)(n + 1) = 0.
     \]
   - Therefore, \( n \leq 3 \).

6. **Check all cases for \( m, n \leq 3 \)**:
   - For \( m, n \leq 3 \), we check all pairs:
     - \( (m, n) = (1, 1) \):
       \[
       1^2 - 1 = 0 \mid 1 + 1^2 = 2 \quad \text{(false)}.
       \]
     - \( (m, n) = (1, 2) \):
       \[
       1^2 - 2 = -1 \mid 1 + 2^2 = 5 \quad \text{(false)}.
       \]
     - \( (m, n) = (2, 1) \):
       \[
       2^2 - 1 = 3 \mid 2 + 1^2 = 3 \quad \text{(true)}.
       \]
     - \( (m, n) = (3, 2) \):
       \[
       3^2 - 2 = 7 \mid 3 + 2^2 = 7 \quad \text{(true)}.
       \]
     - \( (m, n) = (2, 3) \):
       \[
       2^2 - 3 = 1 \mid 2 + 3^2 = 11 \quad \text{(false)}.
       \]
     - \( (m, n) = (2, 2) \):
       \[
       2^2 - 2 = 2 \mid 2 + 2^2 = 6 \quad \text{(false)}.
       \]
     - \( (m, n) = (3, 3) \):
       \[
       3^2 - 3 = 6 \mid 3 + 3^2 = 12 \quad \text{(false)}.
       \]

Conclusion:
The pairs that satisfy both conditions are \( (m, n) = (2, 1) \) and \( (m, n) = (3, 2) \).

The final answer is \(\boxed{(2, 1), (3, 2)}\).