12. On the hyperbola $x y=1$, the point with the abscissa $\frac{n}{n+1}$ is $A_{n}$, and the point with the abscissa $\frac{n+1}{n}$ is $B_{n}\left(n \in \mathbf{N}_{+}\right)$. The point with coordinates $(1,1)$ is denoted as $M, P_{n}\left(x_{n}, y_{n}\right)$ is the circumcenter of $\triangle A_{n} B_{n} M$. Then $x_{1}+x_{2}+\cdots+x_{100}=$ $\qquad$
$12.200 \frac{50}{101}$. $|A_{n} M|=|B_{n} M|$, and $k_{A_{n} B_{n}}=-1$. Therefore, $\triangle M A_{n} B_{n}$ is an isosceles triangle with $A_{n} B_{n}$ as the base, and the slope of the base is -1. Since point $M$ lies on the line $y=x$, the equation of the perpendicular bisector of the base is $y=x$. This gives $x_{n}=y_{n}$.
Thus, the circumcenter $P\left(x_{n}, x_{n}\right)$ lies on the line
$$
y-\frac{2 n+1}{2 n}=\frac{n}{n+1}\left[x-\frac{2 n+1}{2(n+1)}\right]
$$

From this, we get $x_{n}=\frac{(2 n+1)^{2}}{2 n(n+1)}$, i.e.,
$$
\begin{array}{l} 
x_{n}= \frac{(2 n+1)^{2}}{2 n(n+1)}=2+\frac{1}{2 n(n+1)} \\
= 2+\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right) . \\
\text { Therefore, } x_{1}+x_{2}+\cdots+x_{100} \\
= 200+\frac{1}{2}\left(1-\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)+ \\
\cdots+\frac{1}{2}\left(\frac{1}{100}-\frac{1}{101}\right) \\
=200 \frac{50}{101} .
\end{array}
$$