 Two positive integers $m, n$ satisfy the two equations $m^2 + n^2 = 3789$ and $gcd (m, n) + lcm (m, n) = 633$. Compute $m + n$.
1. Let \( \gcd(m, n) = d \). Then we can write \( m = ds \) and \( n = dq \) where \( \gcd(s, q) = 1 \).

2. Substitute \( m = ds \) and \( n = dq \) into the first equation:
   \[
   (ds)^2 + (dq)^2 = 3789
   \]
   \[
   d^2(s^2 + q^2) = 3789
   \]

3. Since \( d^2(s^2 + q^2) = 3789 \), \( d^2 \) must be a divisor of 3789. Factorize 3789:
   \[
   3789 = 3^2 \times 421
   \]
   Thus, \( d^2 \) could be \( 1, 3^2 = 9 \), and hence \( d \) could be \( 1 \) or \( 3 \).

4. Consider the second equation:
   \[
   \gcd(m, n) + \mathrm{lcm}(m, n) = 633
   \]
   Since \( \gcd(m, n) = d \) and \( \mathrm{lcm}(m, n) = \frac{m \cdot n}{\gcd(m, n)} = \frac{ds \cdot dq}{d} = d \cdot s \cdot q \), we have:
   \[
   d + d \cdot s \cdot q = 633
   \]
   \[
   d(1 + s \cdot q) = 633
   \]

5. If \( d = 3 \):
   \[
   3(1 + s \cdot q) = 633
   \]
   \[
   1 + s \cdot q = 211
   \]
   \[
   s \cdot q = 210
   \]

6. Substitute \( d = 3 \) into the equation \( d^2(s^2 + q^2) = 3789 \):
   \[
   9(s^2 + q^2) = 3789
   \]
   \[
   s^2 + q^2 = \frac{3789}{9} = 421
   \]

7. We need to find integers \( s \) and \( q \) such that:
   \[
   s \cdot q = 210
   \]
   \[
   s^2 + q^2 = 421
   \]

8. The pairs \((s, q)\) that satisfy \( s \cdot q = 210 \) are \((1, 210), (2, 105), (3, 70), (5, 42), (6, 35), (7, 30), (10, 21), (14, 15)\).

9. Check which pair satisfies \( s^2 + q^2 = 421 \):
   \[
   s = 14, q = 15 \quad \text{or} \quad s = 15, q = 14
   \]
   \[
   14^2 + 15^2 = 196 + 225 = 421
   \]

10. Therefore, \( m = ds = 3 \times 14 = 42 \) and \( n = dq = 3 \times 15 = 45 \) or vice versa.

11. Finally, compute \( m + n \):
   \[
   m + n = 42 + 45 = 87
   \]

The final answer is \( \boxed{87} \).