Let $m > n$ be positive integers such that $3(3mn - 2)^2 - 2(3m -3n)^2 = 2019$. Find $3m + n$.

1. We start with the given equation:
   \[
   3(3mn - 2)^2 - 2(3m - 3n)^2 = 2019
   \]

2. Expand the squares:
   \[
   3(9m^2n^2 - 12mn + 4) - 2(9m^2 - 18mn + 9n^2)
   \]

3. Distribute the constants:
   \[
   27m^2n^2 - 36mn + 12 - 18m^2 + 36mn - 18n^2
   \]

4. Combine like terms:
   \[
   27m^2n^2 - 18m^2 - 18n^2 + 12 = 2019
   \]

5. Rearrange the equation:
   \[
   27m^2n^2 - 18m^2 - 18n^2 + 12 = 2019
   \]

6. Simplify the equation:
   \[
   27m^2n^2 - 18m^2 - 18n^2 + 12 = 2019
   \]

7. Factor the equation using the method of Simultaneous Substitution and Factoring Technique (SSFT):
   \[
   (9m^2 - 6)(3n^2 - 2) = 2019
   \]

8. Factorize 2019:
   \[
   2019 = 3 \times 673
   \]

9. Set up the equations:
   \[
   9m^2 - 6 = 2019 \quad \text{and} \quad 3n^2 - 2 = 1
   \]

10. Solve for \(m\) and \(n\):
    \[
    9m^2 - 6 = 2019 \implies 9m^2 = 2025 \implies m^2 = 225 \implies m = 15
    \]
    \[
    3n^2 - 2 = 1 \implies 3n^2 = 3 \implies n^2 = 1 \implies n = 1
    \]

11. Calculate \(3m + n\):
    \[
    3m + n = 3 \times 15 + 1 = 45 + 1 = 46
    \]

The final answer is \(\boxed{46}\).