【Analysis】If $A, B, C$ are three rational points on a circle,
then the midpoint $D$ of $AB$ is a rational point, the slope of $AB$ is a rational number or infinite, so the equation of the perpendicular bisector of $AB$ is a linear equation with rational coefficients. Similarly, the equation of the perpendicular bisector of $BC$ is also a linear equation with rational coefficients. Therefore, the intersection of the perpendicular bisector of $AB$ and the perpendicular bisector of $BC$ is a rational point, i.e., the center of the circle is a rational point, which does not meet the condition.

Two examples of rational points: $(0,0)$ and $(1,0)$, the center of the circle is $\left(\frac{1}{2}, \sqrt{2}\right)$, which satisfies the condition.

Therefore, at most two rational points can exist on a circle such that the center of the circle is not a rational point.