Find all pairs $(m, n)$ of positive integers such that $m^2 + 2n^2 = 3(m + 2n)$
1. Start with the given equation:
   \[
   m^2 + 2n^2 = 3(m + 2n)
   \]

2. Rearrange the equation to bring all terms to one side:
   \[
   m^2 + 2n^2 - 3m - 6n = 0
   \]

3. Complete the square for both \(m\) and \(n\):
   \[
   m^2 - 3m + 2n^2 - 6n = 0
   \]
   \[
   (m - \frac{3}{2})^2 - \frac{9}{4} + 2(n - \frac{3}{2})^2 - \frac{18}{4} = 0
   \]
   \[
   (m - \frac{3}{2})^2 + 2(n - \frac{3}{2})^2 = \frac{27}{4}
   \]

4. Multiply through by 4 to clear the fraction:
   \[
   4(m - \frac{3}{2})^2 + 8(n - \frac{3}{2})^2 = 27
   \]
   \[
   (2m - 3)^2 + 2(2n - 3)^2 = 27
   \]

5. Now, we need to find integer solutions to the equation:
   \[
   (2m - 3)^2 + 2(2n - 3)^2 = 27
   \]

6. Consider possible values for \(2m - 3\) and \(2n - 3\) that satisfy the equation. We test the following pairs:
   - \(2m - 3 = 3\) and \(2n - 3 = 3\):
     \[
     (2m - 3)^2 = 3^2 = 9
     \]
     \[
     2(2n - 3)^2 = 2 \cdot 3^2 = 18
     \]
     \[
     9 + 18 = 27
     \]
     Solving for \(m\) and \(n\):
     \[
     2m - 3 = 3 \implies 2m = 6 \implies m = 3
     \]
     \[
     2n - 3 = 3 \implies 2n = 6 \implies n = 3
     \]
     This gives the pair \((m, n) = (3, 3)\).

   - \(2m - 3 = 5\) and \(2n - 3 = 1\):
     \[
     (2m - 3)^2 = 5^2 = 25
     \]
     \[
     2(2n - 3)^2 = 2 \cdot 1^2 = 2
     \]
     \[
     25 + 2 = 27
     \]
     Solving for \(m\) and \(n\):
     \[
     2m - 3 = 5 \implies 2m = 8 \implies m = 4
     \]
     \[
     2n - 3 = 1 \implies 2n = 4 \implies n = 2
     \]
     This gives the pair \((m, n) = (4, 2)\).

7. Therefore, the pairs \((m, n)\) that satisfy the equation are:
   \[
   \boxed{(m, n) = (3, 3), (4, 2)}
   \]