jmc

After some trial and error, we obtain the two lists $\{4,5,7,8,9\}$ and $\{3,6,7,8,9\}$.

Why are these the only two?

If the largest number of the five integers was 8, then the largest that the sum could be would be $8+7+6+5+4=30$, which is too small. This tells us that we must include one 9 in the list. (We cannot include any number larger than 9, since each number must be a single-digit number.)

Therefore, the sum of the remaining four numbers is $33-9=24$.

If the largest of the four remaining numbers is 7, then their largest possible sum would be $7+6+5+4=22$, which is too small. Therefore, we also need to include an 8 in the list.

Thus, the sum of the remaining three numbers is $24-8=16$.

If the largest of the three remaining numbers is 6, then their largest possible sum would be $6+5+4=15$, which is too small. Therefore, we also need to include an 7 in the list.

Thus, the sum of the remaining two numbers is $16-7=9$.

This tells us that we need two different positive integers, each less than 7, that add to 9. These must be 3 and 6 or 4 and 5.

This gives us the two lists above, and shows that they are the only two such lists. The answer is $\boxed{2}$.