Find all pairs of integers $(x, y)$ such that ths sum of the fractions $\frac{19}{x}$ and $\frac{96}{y}$ would be equal to their product.
1. We start with the given equation:
   \[
   \frac{19}{x} + \frac{96}{y} = \frac{19 \cdot 96}{xy}
   \]

2. To simplify, we multiply both sides by \(xy\):
   \[
   19y + 96x = 19 \cdot 96
   \]

3. Simplify the right-hand side:
   \[
   19y + 96x = 1824
   \]

4. Rearrange the equation to isolate \(y\):
   \[
   19y = 1824 - 96x
   \]

5. Since \(19y\) must be divisible by 19, \(1824 - 96x\) must also be divisible by 19. We check the divisibility:
   \[
   1824 \equiv 0 \pmod{19}
   \]
   \[
   96x \equiv 0 \pmod{19}
   \]

6. Simplify \(96 \pmod{19}\):
   \[
   96 \equiv 1 \pmod{19}
   \]

7. Therefore, the equation becomes:
   \[
   x \equiv 0 \pmod{19}
   \]

8. Let \(x = 19m\) for some integer \(m\). Substitute \(x\) back into the equation:
   \[
   19y + 96(19m) = 1824
   \]
   \[
   19y + 1824m = 1824
   \]

9. Simplify by dividing the entire equation by 19:
   \[
   y + 96m = 96
   \]
   \[
   y = 96 - 96m
   \]

10. Therefore, the pairs \((x, y)\) are:
    \[
    (x, y) = (19m, 96 - 96m)
    \]

The final answer is \( \boxed{ (19m, 96 - 96m) } \) for any integer \(m\).