Let $a, b, c$ be integers not all the same with $a, b, c\ge 4$ that satisfy $$4abc = (a + 3) (b + 3) (c + 3).$$
Find the numerical value of $a + b + c$.
1. Given the equation:
   \[
   4abc = (a + 3)(b + 3)(c + 3)
   \]
   where \(a, b, c\) are integers not all the same and \(a, b, c \geq 4\).

2. Divide both sides by \(abc\):
   \[
   4 = \left(1 + \frac{3}{a}\right)\left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)
   \]

3. Consider the case where none of \(a, b, c\) is equal to 4. The greatest possible value of \(\left(1 + \frac{3}{a}\right)\left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)\) with \(a, b, c \geq 4\) and not all equal would be:
   \[
   \left(1 + \frac{3}{5}\right)\left(1 + \frac{3}{5}\right)\left(1 + \frac{3}{6}\right) = \frac{8}{5} \cdot \frac{8}{5} \cdot \frac{9}{6} = \frac{64}{25} \cdot \frac{3}{2} = \frac{192}{50} = \frac{96}{25} < 4
   \]
   Therefore, at least one of \(a, b, c\) must be equal to 4.

4. Without loss of generality, let \(a = 4\). Substitute \(a = 4\) into the equation:
   \[
   4 = \left(1 + \frac{3}{4}\right)\left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)
   \]
   Simplify:
   \[
   4 = \frac{7}{4}\left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)
   \]
   Multiply both sides by 4:
   \[
   16 = 7\left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)
   \]
   Divide both sides by 7:
   \[
   \frac{16}{7} = \left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)
   \]

5. To satisfy \(\frac{16}{7} = \left(1 + \frac{3}{b}\right)\left(1 + \frac{3}{c}\right)\), at least one of \(b\) or \(c\) must be divisible by 7. Assume \(b = 7\). Substitute \(b = 7\) into the equation:
   \[
   \frac{16}{7} = \left(1 + \frac{3}{7}\right)\left(1 + \frac{3}{c}\right)
   \]
   Simplify:
   \[
   \frac{16}{7} = \frac{10}{7}\left(1 + \frac{3}{c}\right)
   \]
   Multiply both sides by 7:
   \[
   16 = 10\left(1 + \frac{3}{c}\right)
   \]
   Divide both sides by 10:
   \[
   \frac{8}{5} = 1 + \frac{3}{c}
   \]
   Subtract 1 from both sides:
   \[
   \frac{3}{5} = \frac{3}{c}
   \]
   Therefore:
   \[
   c = 5
   \]

6. Thus, we have \(a = 4\), \(b = 7\), and \(c = 5\). Therefore:
   \[
   a + b + c = 4 + 7 + 5 = 16
   \]

The final answer is \(\boxed{16}\).