One, (50 points) The product of all elements in a finite set $S$ is called the "product number" of the set $S$. Given the set $M=$ $\left\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \cdots, \frac{1}{99}, \frac{1}{100}\right\}$, determine the sum of the "product numbers" of all even-numbered (2, 4, $\cdots$, 98) element subsets of $M$.
Let the set $M=\left\{\frac{1}{2}, \frac{1}{3},-\frac{1}{4}, \cdots, \frac{i}{99}, \frac{1}{150}\right\}, M$ contains 99 different elements. Let $G$ be the sum of the "product numbers" of all even-numbered element subsets of $M$, and $H$ be the sum of the "product numbers" of all odd-numbered element subsets of $M$. It is easy to see that
$$
\begin{aligned}
G+H= & \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \times \cdots \\
& \times\left(1+\frac{1}{99}\right)\left(1+\frac{1}{100}\right)-1 \\
= & \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{100}{99} \times \frac{101}{100}-1 \\
= & \frac{101}{2}-1=\frac{99}{2} . \\
G-H= & \left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \times \cdots \\
& \times\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)-1 \\
= & \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{98}{99} \times \frac{99}{100}-1 \\
= & \frac{1}{100}-1=-\frac{99}{100} .
\end{aligned}
$$

Adding these, we get $G=\frac{1}{2}\left(\frac{99}{2}-\frac{99}{100}\right)=\frac{99 \times 49}{200}=24.255$. Therefore, the sum of the "product numbers" of all even-numbered element subsets of $M$ is
24.255.