jmc

We have the smallest stack, which has a height of $94\times 4$ inches. Now when we change the height of one of the bricks, we either add $0$ inches, $6$ inches, or $15$ inches to the height. Now all we need to do is to find the different change values we can get from $94$ $0$'s, $6$'s, and $15$'s. Because $0$, $6$, and $15$ are all multiples of $3$, the change will always be a multiple of $3$, so we just need to find the number of changes we can get from $0$'s, $2$'s, and $5$'s. From here, we count what we can get: $$0,2=2,4=2+2,5=5,6=2+2+2,7=5+2,8=2+2+2+2,9=5+2+2,\dots$$ It seems we can get every integer greater or equal to four; we can easily deduce this by considering parity or using the Chicken McNugget Theorem, which says that the greatest number that cannot be expressed in the form of $2m+5n$ for $m,n$ being positive integers is $5\times 2-5-2=3$. But we also have a maximum change ($94\times 5$), so that will have to stop somewhere. To find the gaps, we can work backwards as well. From the maximum change, we can subtract either $0$'s, $3$'s, or $5$'s. The maximum we can't get is $5\times 3-5-3=7$, so the numbers $94\times 5-8$ and below, except $3$ and $1$, work. Now there might be ones that we haven't counted yet, so we check all numbers between $94\times 5-8$ and $94\times 5$. $94\times 5-7$ obviously doesn't work, $94\times 5-6$ does since 6 is a multiple of 3, $94\times 5-5$ does because it is a multiple of $5$ (and $3$), $94\times 5-4$ doesn't since $4$ is not divisible by $5$ or $3$, $94\times 5-3$ does since $3=3$, and $94\times 5-2$ and $94\times 5-1$ don't, and $94\times 5$ does. Thus the numbers $0$, $2$, $4$ all the way to $94\times 5-8$, $94\times 5-6$, $94\times 5-5$, $94\times 5-3$, and $94\times 5$ work. That's $2+(94\times 5-8-4+1)+4=\boxed{465}$ numbers.