10. (This question is worth 20 points) Given that $f(x)$ is an odd function on $\mathbf{R}$, $f(1)=1$, and for any $x<0$, $f\left(\frac{x}{x-1}\right)=x f(x)$. Find the value of $f(1) f\left(\frac{1}{100}\right)+f\left(\frac{1}{2}\right) f\left(\frac{1}{99}\right)+f\left(\frac{1}{3}\right) f\left(\frac{1}{98}\right)+\cdots+f\left(\frac{1}{50}\right) f\left(\frac{1}{51}\right)$.
Solution: Let $a_{n}=f\left(\frac{1}{n}\right)(n=1,2,3, \cdots)$, then $a_{1}=f(1)=1$.
In $f\left(\frac{x}{x-1}\right)=x f(x)$, take $x=-\frac{1}{k}\left(k \in \mathbf{N}^{*}\right)$, noting that $\frac{x}{x-1}=\frac{-\frac{1}{k}}{-\frac{1}{k}-1}=\frac{1}{k+1}$, and $f(x)$ is an odd function, we have
$$
f\left(\frac{1}{k+1}\right)=-\frac{1}{k} \cdot f\left(-\frac{1}{k}\right)=\frac{1}{k} \cdot f\left(\frac{1}{k}\right),
$$

i.e., $\frac{a_{k+1}}{a_{k}}=\frac{1}{k}$. Thus, $a_{n}=a_{1} \cdot \prod_{k=1}^{n-1} \frac{a_{k+1}}{a_{k}}=\prod_{k=1}^{n-1} \frac{1}{k}=\frac{1}{(n-1)!}$.
Therefore,
$$
\begin{array}{c}
\sum_{i=1}^{50} a_{i} a_{101-i}=\sum_{i=1}^{50} \frac{1}{(i-1)!\cdot(100-i)!}=\sum_{i=0}^{49} \frac{1}{i!\cdot(99-i)!} \\
=\frac{1}{99!} \cdot \sum_{i=0}^{49} \mathrm{C}_{99}^{i}=\frac{1}{99!} \cdot \sum_{i=0}^{49} \frac{1}{2}\left(\mathrm{C}_{99}^{i}+\mathrm{C}_{99}^{99-i}\right)=\frac{1}{99!} \times \frac{1}{2} \times 2^{99}=\frac{2^{98}}{99!} .
\end{array}
$$
20 points