12. Given the function $f(x)$ satisfies: $f(2-x)=2-f(x)(x \in \mathbf{R})$, the intersection points of the graphs of $y=\frac{x}{x-1}$ and $y=f(x)$ are $\left(x_{i}, y_{i}\right)(i=1,2, \cdots, n)$, then $\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=$ $\qquad$ .
From $f(2-x)+f(x)=2$ we know that the graph of $f(x)$ is symmetric about the point $(1,1)$, and the graph of $y=\frac{x}{x-1}=1+\frac{1}{x-1}$ is also symmetric about the point $(1,1)$. Therefore, the $n$ intersection points $\left(x_{i}, y_{i}\right)(i=1,2, \cdots, n)$ are symmetric about the point $(1,1)$. Thus, $\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=2 n$.