Positive integers m and n are both greater than 50, have a least common multiple equal to 480, and have a
greatest common divisor equal to 12. Find m + n.

1. We start with the given information:
   - Least Common Multiple (LCM) of \( m \) and \( n \) is 480.
   - Greatest Common Divisor (GCD) of \( m \) and \( n \) is 12.
   - Both \( m \) and \( n \) are greater than 50.

2. We use the relationship between LCM and GCD:
   \[
   \text{LCM}(m, n) \times \text{GCD}(m, n) = m \times n
   \]
   Substituting the given values:
   \[
   480 \times 12 = m \times n
   \]
   \[
   5760 = m \times n
   \]

3. We need to find pairs \((m, n)\) such that \( m \times n = 5760 \) and both \( m \) and \( n \) are greater than 50. Additionally, \( \text{GCD}(m, n) = 12 \).

4. We express \( m \) and \( n \) in terms of their GCD:
   \[
   m = 12a \quad \text{and} \quad n = 12b
   \]
   where \( \text{GCD}(a, b) = 1 \). Substituting these into the product equation:
   \[
   (12a) \times (12b) = 5760
   \]
   \[
   144ab = 5760
   \]
   \[
   ab = \frac{5760}{144}
   \]
   \[
   ab = 40
   \]

5. We need to find pairs \((a, b)\) such that \( ab = 40 \) and \( \text{GCD}(a, b) = 1 \). The pairs \((a, b)\) that satisfy this are:
   \[
   (1, 40), (2, 20), (4, 10), (5, 8)
   \]

6. We now convert these pairs back to \( m \) and \( n \):
   \[
   (m, n) = (12a, 12b)
   \]
   - For \((1, 40)\): \((m, n) = (12 \times 1, 12 \times 40) = (12, 480)\) (not valid since both must be > 50)
   - For \((2, 20)\): \((m, n) = (12 \times 2, 12 \times 20) = (24, 240)\) (not valid since both must be > 50)
   - For \((4, 10)\): \((m, n) = (12 \times 4, 12 \times 10) = (48, 120)\) (not valid since both must be > 50)
   - For \((5, 8)\): \((m, n) = (12 \times 5, 12 \times 8) = (60, 96)\) (valid since both are > 50)

7. The only valid pair is \((60, 96)\). Therefore, the sum \( m + n \) is:
   \[
   60 + 96 = 156
   \]

The final answer is \(\boxed{156}\).