1.35 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 have when all 94 bricks are used?
[Solution] First note that if 5 bricks each contribute a height of 10 to the tower, then when 3 of them are laid flat, each contributing a height of 4, and the other two are stood up, each contributing a height of 19, the height of the tower remains unchanged. Therefore, when calculating the different heights of the tower, we can always assume that the number of bricks contributing 10 is less than 5.

Let the number of bricks contributing 10 and 19 to the height of the tower be $x, y$, respectively, then the number of bricks contributing 4 is $94-x-y$. Thus, the height of the tower is
$$
4(94-x-y)+10 x+19 y=376+6 x+15 y \text {, }
$$

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

Next, we prove that the 465 values given by (1) are all distinct. If not, suppose there are different pairs $(x, y),(u, v)$ corresponding to the same tower height, i.e.,
$$
376+6 x+15 y=376+6 u+15 v .
$$

From (2), we get
$$
2(x-u)=5(v-y) .
$$
Equation (3) indicates that $5 \mid(x-u)$. Since $0 \leqslant x \leqslant 4,0 \leqslant u \leqslant 4$, it follows that $-4 \leqslant x-u \leqslant$
4. Therefore, it must be that $x=u$. From this and (3), we also get $v=y$, a contradiction.
In summary, the height of the tower has a total of 465 different values.