9. Let the function
$$
\begin{array}{l}
f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}, \\
S_{n}=\sum_{i=1}^{n-1} f\left(\frac{i}{n}\right)\left(n \geqslant 2, n \in \mathbf{N}_{+}\right) .
\end{array}
$$

Then $S_{n}=$
9. $\frac{n-1}{2}$.

When $x_{1}+x_{2}=1$,
$$
f\left(x_{1}\right)+f\left(x_{2}\right)=1+\log _{2} \frac{x_{1} x_{2}}{\left(1-x_{1}\right)\left(1-x_{2}\right)}=1 \text {. }
$$

Then $2 S_{n}=\sum_{i=1}^{n-1}\left[f\left(\frac{i}{n}\right)+f\left(\frac{n-i}{n}\right)\right]=n-1$
$$
\Rightarrow S_{n}=\frac{n-1}{2} \text {. }
$$