17. Happy Sums (from 8th grade, 4 points). In the "Happy Sum" lottery, there are $N$ balls numbered from 1 to $N$. During the main draw, 10 balls are randomly drawn. During the additional draw, 8 balls are randomly selected from the same set of balls. The sum of the numbers on the drawn balls in each draw is announced as the happy sum, and players who predicted this sum win a prize.

Can it be that events $A$ "the happy sum in the main draw is 63" and $B$ "the happy sum in the additional draw is 44" are equally likely? If yes, under what condition?
Solution. In the main draw, there are $C_{N}^{10}$ possible combinations, and in the additional draw, there are $C_{N}^{8}$ combinations. Let $S_{63,10}$ be the number of combinations in the first draw where the sum is 63. The addends are different natural numbers from 1 to $N$. If the smallest number is 2 or more, then the sum of all ten addends is no less than

$$
2+3+\ldots+11=\frac{2+11}{2} \cdot 10=13 \cdot 5=65
$$

Thus, the smallest number is 1. By removing one from each addend, i.e., subtracting 10 from the sum, we find that the remaining sum of 53 is obtained by nine different natural addends, each not exceeding the number $N$. In other words, the number of ways to get a sum of 63 with ten addends is the same as the number of ways to get a sum of 53 with nine addends: $S_{63,10}=S_{53,9}$.

Reasoning similarly, we get: $S_{63,10}=S_{53,9}=S_{44,8}$. Now we can find the ratio of probabilities:

$$
\frac{\mathrm{P}(B)}{\mathrm{P}(A)}=\frac{S_{44,8} \cdot C_{N}^{10}}{C_{N}^{8} \cdot S_{63,10}}=\frac{C_{N}^{10}}{C_{N}^{8}}=\frac{8!(N-8)!}{10!(N-10)!}=\frac{(N-8)(N-9)}{90}
$$

Let's find for which $N$ the obtained fraction equals one:

$$
\frac{(N-8)(N-9)}{90}=1 ; N^{2}-17 N-18=0, \text { from which } N=18
$$

Answer: This is possible if there are 18 balls in the lottery.