For $B_{n}\left(n \in \mathbf{N}^{*}\right)$. Let the point with coordinates $(1,1)$ be $M, P_{n}\left(x_{n}, y_{n}\right)$ is the circumcenter of $\triangle A_{n} B_{n} M$, find the value of $x_{1}+x_{2}+\cdots+x_{100}$.
Given $A_{n}\left(\frac{n}{n+1}, \frac{n+1}{n}\right), B_{n}\left(\frac{n+1}{n}, \frac{n}{n+1}\right)$, we have $\left|A_{n} M\right|=\left|B_{n} M\right|$, 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 $M$ lies on the line $y=x$, the perpendicular bisector of the base has the equation $y=x$, which implies $x_{n}=y_{n}$.

The midpoint of $A_{n} M$ is $E\left(\frac{2 n+1}{2(n+1)}, \frac{2 n+1}{2 n}\right)$, and $k_{A_{n} M}=\frac{\frac{n+1}{n}-1}{\frac{n}{n+1}-1}=-\frac{n+1}{n}$. Therefore, 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 which we get $x_{n}=$ $\frac{(2 n+1)^{2}}{2 n(n+1)}$, i.e.,
$$
x_{n}=\frac{(2 n+1)^{2}}{2 n(n+1)}=2+\frac{1}{2 n(n+1)}=2+\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right) .
$$

Thus, $x_{1}+x_{2}+\cdots+x_{100}=200+\frac{1}{2}\left(1-\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+$
$$
\frac{1}{2}\left(\frac{1}{100}-\frac{1}{101}\right)=200 \frac{50}{101} .
$$