24. A bag contains 999 balls, marked with the 999 numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{1000}$ respectively. Each ball is either red or blue, and the number of red balls is a positive even number. Let $S$ denote the product of the numbers on all red balls. Find the sum of all possible values of $S$.
A bag contains 999 balls, marked with the 999 numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{1000}$ respectively. Each ball is either red or blue, and the number of red balls is a positive even number. Let $S$ denote the product of the numbers on all red balls. Find the sum of all possible values of $S$.
24. (Cancelled)
24. (This question was cancelled in the live paper. We present below a solution to find the sum of all possible values of \( S \) counting multiplicities, i.e., if a possible value of \( S \) is attained in two different situations, we shall sum that value twice.)
Let's also take into consideration the case where the number of red balls is a positive odd number, and in this case let \( T \) be the product of the numbers on all red balls. Let \( G \) be the sum of all possible values of \( S \) (counting multiplicities) and \( H \) be the sum of all possible values of \( T \) (counting multiplicities). Note that
\[
G+H=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{1000}\right)-1=\frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{1001}{1000}-1=\frac{1001}{2}-1=\frac{999}{2}
\]
since each possible value of \( S \) or \( T \) arises from a choice of whether each ball is red or blue, and the term -1 in the end eliminates the case where all balls are blue which is not allowed by the question. In a similar manner, we have
\[
G-H=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{1000}\right)-1=\frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{999}{1000}-1=\frac{1}{1000}-1=-\frac{999}{1000}
\]
since each possible value of \( T \) (corresponding to a choice of an odd number of red balls) is evaluated -1 time in the above product while each possible value of \( S \) is evaluated +1 time. It follows that
\[
G=\frac{1}{2}\left(\frac{999}{2}-\frac{999}{1000}\right)=\frac{1}{2}\left(\frac{499500-999}{1000}\right)=\frac{498501}{2000}.
\]