Find all positive integers $ a$ and $ b$ for which 

\[ \left \lfloor \frac{a^2}{b} \right \rfloor \plus{} \left \lfloor \frac{b^2}{a} \right \rfloor \equal{} \left \lfloor \frac{a^2 \plus{} b^2}{ab} \right \rfloor \plus{} ab.\]
1. **Claim and Initial Setup**:
   We claim that all ordered pairs of integers \((a, b)\) that satisfy the given equation are of the form \((k, k^2+1)\) or \((k^2+1, k)\) for \(k \in \mathbb{Z}^{+}\).

2. **Substitution and Simplification**:
   Let \(a = b^2 + 1\). If \(b > 1\), then we have:
   \[
   \text{LHS} = \left\lfloor \frac{a^2}{b} \right\rfloor + \left\lfloor \frac{b^2}{a} \right\rfloor = \left\lfloor \frac{(b^2+1)^2}{b} \right\rfloor + \left\lfloor \frac{b^2}{b^2+1} \right\rfloor
   \]
   Simplifying each term:
   \[
   \left\lfloor \frac{(b^2+1)^2}{b} \right\rfloor = \left\lfloor b^3 + 2b + \frac{1}{b} \right\rfloor = b^3 + 2b
   \]
   and
   \[
   \left\lfloor \frac{b^2}{b^2+1} \right\rfloor = 0
   \]
   Thus,
   \[
   \text{LHS} = b^3 + 2b
   \]

3. **Right-Hand Side Calculation**:
   \[
   \text{RHS} = \left\lfloor \frac{a^2 + b^2}{ab} \right\rfloor + ab = \left\lfloor \frac{(b^2+1)^2 + b^2}{b(b^2+1)} \right\rfloor + b(b^2+1)
   \]
   Simplifying the fraction inside the floor function:
   \[
   \frac{(b^2+1)^2 + b^2}{b(b^2+1)} = \frac{b^4 + 2b^2 + 1 + b^2}{b(b^2+1)} = \frac{b^4 + 3b^2 + 1}{b^3 + b}
   \]
   For large \(b\), this fraction approximates to \(b + \frac{3}{b} + \frac{1}{b^3}\), so:
   \[
   \left\lfloor b + \frac{3}{b} + \frac{1}{b^3} \right\rfloor = b
   \]
   Thus,
   \[
   \text{RHS} = b + b^3 + b = b^3 + 2b
   \]

4. **Verification**:
   Since \(\text{LHS} = \text{RHS}\), the pair \((a, b) = (b^2 + 1, b)\) is a solution. By symmetry, \((b, a) = (b, b^2 + 1)\) is also a solution.

5. **Exclusion of Other Solutions**:
   Assume \(b \ge a\). Let \(f(b) = \frac{a^2}{b} + \frac{b^2}{a} - \left(\frac{a}{b} + \frac{b}{a}\right) - ab\) and \(g(b) = \frac{a^2}{b} + \frac{b^2}{a} - \frac{a}{b} - \frac{b}{a} - ab\). We need to show that \(f(b) \neq 0\) for other values of \(b\).

6. **Bounding \(g(b)\)**:
   \[
   g(a) = \frac{1}{a} \left(b - \frac{a^2 + 1}{2}\right)^2 = \frac{1}{4a} \left(\frac{a^2 + 1}{2}\right)^2 = \frac{a^2 - a}{b}
   \]
   If \(b \ge a^2 + 2\), then:
   \[
   g(b) \ge a + \frac{2}{a} + \frac{a^2 - a}{a^2 + 2} > a
   \]
   Hence, \(f(b) > a - 3\).

7. **Checking Small Values**:
   For \(a = 1\) and \(a = 2\), we manually check the values of \(b\) and find that the only solutions are \(b = 2\) for \(a = 1\) and no other solutions for \(a = 2\).

8. **Conclusion**:
   Hence, the only solutions are of the form \((k, k^2 + 1)\) or \((k^2 + 1, k)\) for \(k \in \mathbb{Z}^{+}\).

\(\blacksquare\)

The final answer is \( \boxed{ (k, k^2 + 1) } \) or \((k^2 + 1, k)\) for \(k \in \mathbb{Z}^{+}\).