Find the solutions of positive integers for the system $xy + x + y = 71$ and $x^2y + xy^2 = 880$.
1. Given the system of equations:
   \[
   xy + x + y = 71
   \]
   \[
   x^2y + xy^2 = 880
   \]

2. Let \( a = xy \) and \( b = x + y \). Substituting these into the equations, we get:
   \[
   a + b = 71
   \]
   \[
   x^2y + xy^2 = xy(x + y) = a \cdot b = 880
   \]

3. We now have the system:
   \[
   a + b = 71
   \]
   \[
   ab = 880
   \]

4. To find \(a\) and \(b\), we solve the quadratic equation formed by these relationships. Using Vieta's formulas, we know that \(a\) and \(b\) are the roots of the quadratic equation:
   \[
   t^2 - (a+b)t + ab = 0
   \]
   Substituting \(a + b = 71\) and \(ab = 880\), we get:
   \[
   t^2 - 71t + 880 = 0
   \]

5. Solving this quadratic equation using the quadratic formula \( t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
   \[
   t = \frac{71 \pm \sqrt{71^2 - 4 \cdot 880}}{2}
   \]
   \[
   t = \frac{71 \pm \sqrt{5041 - 3520}}{2}
   \]
   \[
   t = \frac{71 \pm \sqrt{1521}}{2}
   \]
   \[
   t = \frac{71 \pm 39}{2}
   \]

6. This gives us two solutions for \(t\):
   \[
   t = \frac{71 + 39}{2} = 55
   \]
   \[
   t = \frac{71 - 39}{2} = 16
   \]

7. Therefore, \(a = 16\) and \(b = 55\), or \(a = 55\) and \(b = 16\).

8. Now, we need to find \(x\) and \(y\) such that \(xy = a\) and \(x + y = b\). We consider both cases:

   - For \(a = 16\) and \(b = 55\):
     \[
     t^2 - 55t + 16 = 0
     \]
     The discriminant of this quadratic is:
     \[
     \Delta = 55^2 - 4 \cdot 1 \cdot 16 = 3025 - 64 = 2961
     \]
     Since the discriminant is not a perfect square, this quadratic does not have integer roots.

   - For \(a = 55\) and \(b = 16\):
     \[
     t^2 - 16t + 55 = 0
     \]
     The discriminant of this quadratic is:
     \[
     \Delta = 16^2 - 4 \cdot 1 \cdot 55 = 256 - 220 = 36
     \]
     Since the discriminant is a perfect square, this quadratic has integer roots:
     \[
     t = \frac{16 \pm \sqrt{36}}{2} = \frac{16 \pm 6}{2}
     \]
     \[
     t = \frac{22}{2} = 11 \quad \text{or} \quad t = \frac{10}{2} = 5
     \]

9. Therefore, the solutions for \(x\) and \(y\) are:
   \[
   (x, y) = (11, 5) \quad \text{or} \quad (5, 11)
   \]

The final answer is \(\boxed{(x, y) = (11, 5) \text{ or } (5, 11)}\).