10. (20 points) Given that $f(x)$ is an odd function on $\mathbf{R}$, $f(1)=1$, and for any $x<0$,
$$
f\left(\frac{x}{x-1}\right)=x f(x) \text {. }
$$

Find the value of $\sum_{i=1}^{50} f\left(\frac{1}{i}\right) f\left(\frac{1}{101-i}\right)$.
10. Let $a_{n}=f\left(\frac{1}{n}\right)(n=1,2, \cdots)$.

Then $a_{1}=f(1)=1$.
In $f\left(\frac{x}{x-1}\right)=x f(x)$, take
$$
x=-\frac{1}{k}\left(k \in \mathbf{Z}_{+}\right) \text {. }
$$

Notice that, $\frac{x}{x-1}=\frac{-\frac{1}{k}}{-\frac{1}{k}-1}=\frac{1}{k+1}$, and $f(x)$
is an odd function.
$$
\begin{array}{l}
\text { Hence } f\left(\frac{1}{k+1}\right)=-\frac{1}{k} f\left(-\frac{1}{k}\right)=\frac{1}{k} f\left(\frac{1}{k}\right) \\
\Rightarrow \frac{a_{k+1}}{a_{k}}=\frac{1}{k} \\
\Rightarrow a_{n}=a_{1} \prod_{k=1}^{n-1} \frac{a_{k+1}}{a_{k}}=\prod_{k=1}^{n-1} \frac{1}{k}=\frac{1}{(n-1)!} \text {. } \\
\text { Then } \sum_{i=1}^{50} a_{i} a_{101-i}=\sum_{i=1}^{50} \frac{1}{((i-1)!)((100-i)!)} \\
=\sum_{i=0}^{49} \frac{1}{(i!)((99-i)!)}=\frac{1}{99!} \sum_{i=0}^{49} \mathrm{C}_{99}^{i} \\
=\frac{1}{99!} \sum_{i=0}^{49} \frac{1}{2}\left(\mathrm{C}_{99}^{i}+\mathrm{C}_{99}^{99-i}\right) \\
=\frac{1}{99!} \times \frac{1}{2} \times 2^{99}=\frac{2^{98}}{99!} . \\
\end{array}
$$