Example 2: Using 94 bricks of size $4 \times 10 \times 19$, stack them one on top of the other to form a tower 94 bricks high. Each brick can be placed in any orientation, providing heights of 4, 10, or 19. How many different values can the height of the tower take after all 14 bricks are placed?
(1994 12th American Invitational Mathematics Examination)
First, note that if 5 bricks each provide a height of 10 to the tower, then placing 3 of them flat, each providing a height of 4, and the other two standing up, each providing a height of 19, the total height provided by these 5 bricks remains 50. Therefore, when calculating all possible different heights of the tower, we can always assume that the number of bricks providing a height of 10 is less than 5.

Let the number of bricks contributing a height of 10 and 19 be $x$ and $y$, respectively. Then the number of bricks contributing a height of 4 is $94 - x - y$, and the height of the tower is
$$
h = 4(94 - x - y) + 10x + 19y = 376 + 6x + 15y.
$$

where $0 \leqslant x \leqslant 4, 0 \leqslant y \leqslant 94 - x$, thus there are $95 + 94 + 93 + 92 + 91 = 465$ different non-negative integer pairs $(x, y)$, so the height of the tower can have at most 465 different values.

Next, we prove that the 465 values given by (1) are all different. If not, suppose there are different pairs $(x, y)$ and $(u, v)$ corresponding to the same tower height, i.e., $376 + 6x + 15y = 376 + 6u + 15v$.
Then we have $2(x - u) = 5(v - y)$.
This indicates that $5 \mid x - u$, because $0 \leqslant x \leqslant 4, 0 \leqslant u \leqslant 4$, hence $-4 \leqslant x - u \leqslant 4$, thus it must be that $x = u$. From (2), we get $y = v$, which is a contradiction.
In summary, the height of the tower has a total of 465 different values.