B. In the Cartesian coordinate system $x O y$, it is known that $O$ is the origin, point $A(10,100), B\left(x_{0}, y_{0}\right)$, where $x_{0} 、 y_{0}$ are integers, and points $O 、 A 、 B$ are not collinear. For all points $B$ that satisfy the above conditions, find the minimum area of $\triangle O A B$.
B. Draw a line parallel to the $x$-axis through point $B$, intersecting line $O A$ at point $C\left(\frac{y_{0}}{10}, y_{0}\right)$.
Thus, $B C=\left|x_{0}-\frac{y_{0}}{10}\right|$.
Since points $O$, $A$, and $B$ are not collinear, we have
$$
\begin{array}{l}
S_{\triangle O A B}=\frac{1}{2} B C \times 100=50\left|x_{0}-\frac{y_{0}}{10}\right| \\
=5\left|10 x_{0}-y_{0}\right| \neq 0 .
\end{array}
$$

Given that $\left|10 x_{0}-y_{0}\right|$ is an integer, then $\left|10 x_{0}-y_{0}\right| \geqslant 1$.
Therefore, $S_{\triangle O A B}=5\left|10 x_{0}-y_{0}\right| \geqslant 5$.
When $x_{0}=1, y_{0}=9$, $S_{\triangle O A B}=5$.
Thus, the minimum area of $\triangle O A B$ is 5.