Problem 1319

Given the equation:
$$\frac{a^2-2}{x} = \frac{b^2-37}{y} = \frac{c^2-41}{z} = a+b+c$$
Let $M = a + b + c$. Since $a, b, c \geq 1$, we have $M \geq 3$.

### Lemma 1: $M \mid 146$
Proof of Lemma 1:
Consider the quantity $T = (a + b + c)(-41ab + 3bc + 38ca)$.
Note that $T \equiv 0 \pmod{M}$.

But,
$$T \equiv -41a^2b + 38a^2c - 41b^2a + 3b^2c + 3c^2b + 38c^2a \pmod{M}$$
Rewriting the terms, we get:
$$T \equiv [(-41a^2b + 82b) + (38a^2c - 76c) + (-41b^2a + 1517a) + (3b^2c - 111c) + (3c^2b - 123b) + (38c^2a - 1558a)] + [41a + 41b + 187c] \pmod{M}$$
Simplifying further:
$$T \equiv 41a + 41b + 187c \equiv 146c \pmod{M}$$
Thus, $M \mid 146c$.

Now, if $\gcd(M, c) = k$, then $k \mid 41$. Since $\gcd(146, 41) = 1$, we have $M \mid 146$.

This completes the proof of Lemma 1. $\blacksquare$

The only factors of $146$ are $1, 2, 73, 146$. Since $M \geq 3$, we have $M = 73$ or $M = 146$.

### Case 1: $M = 73$
We need to find $a, b, c$ such that:
$$a + b + c = 73$$
Given the conditions:
$$\frac{a^2-2}{x} = \frac{b^2-37}{y} = \frac{c^2-41}{z} = 73$$
We can solve for $x, y, z$:
$$x = \frac{a^2-2}{73}, \quad y = \frac{b^2-37}{73}, \quad z = \frac{c^2-41}{73}$$
Checking possible values, we find:
$$a = 32, \quad b = 16, \quad c = 25$$
Thus:
$$x = \frac{32^2-2}{73} = \frac{1024-2}{73} = 14$$
$$y = \frac{16^2-37}{73} = \frac{256-37}{73} = 3$$
$$z = \frac{25^2-41}{73} = \frac{625-41}{73} = 8$$
So, $(a, b, c, x, y, z) = (32, 16, 25, 14, 3, 8)$.

Thus, $S = a + b + c + x + y + z = 32 + 16 + 25 + 14 + 3 + 8 = 98$.

### Case 2: $M = 146$
We need to find $a, b, c$ such that:
$$a + b + c = 146$$
Given the conditions:
$$\frac{a^2-2}{x} = \frac{b^2-37}{y} = \frac{c^2-41}{z} = 146$$
We can solve for $x, y, z$:
$$x = \frac{a^2-2}{146}, \quad y = \frac{b^2-37}{146}, \quad z = \frac{c^2-41}{146}$$
Checking possible values, we find:
$$a = 32, \quad b = 89, \quad c = 25$$
Thus:
$$x = \frac{32^2-2}{146} = \frac{1024-2}{146} = 7$$
$$y = \frac{89^2-37}{146} = \frac{7921-37}{146} = 54$$
$$z = \frac{25^2-41}{146} = \frac{625-41}{146} = 4$$
So, $(a, b, c, x, y, z) = (32, 89, 25, 7, 54, 4)$.

Thus, $S = a + b + c + x + y + z = 32 + 89 + 25 + 7 + 54 + 4 = 211$.

### Conclusion
The sum of all possible values of $S$ is:
$$98 + 211 = \boxed{309}$$