Prove there are at least $2$ lines with the property that each of them divides the plane into $2$ regions with the same number of red and blue points. [duplicate]

At least $3$ of the points must be vertices of their convex hull. Let $p$ be one of these, and let $\ell$ be a line through $p$ that has all of the other $2n-1$ points on one side of it. Without loss of generality we may establish a coordinate system with $p$ at the origin, $\ell$ as the $x$-axis, and all of the other points above the $x$-axis, and we may suppose that $p$ is red. The open half space above $\ell$ has $n$ blue points and $n-1$ red points, so that half space has a surplus of $1$ blue point. Now rotate the coordinate system counterclockwise about the origin $p$. The $x$-axis passes through the other $2n-1$ points one at a time. For $0\le\theta\le\pi$ let $d(\theta)$ be the excess of blue points over red points in the open upper half plane, so that $d(0)=1$, and $d(\pi)=0$. Clearly $d(\theta)$ is piecewise constant, changing only when the $x$-axis hits one of the other $2n-1$ points. Specifically, $d(\theta)$ decreases by $1$ when the $x$-axis hits a blue point and increases by $1$ when it hits a red point.

Let $\theta_0=\min\{\theta\in[0,\pi]:d(\theta)=0\}$. Since $d(0)=1>0$, and $d(\theta)$ can change only in steps of $\pm1$, the $x$-axis at angle $\theta_0$ must pass through $p$ and a blue point $q$ and must therefore be a balanced line.

As noted at the beginning, there must be at least one vertex of the convex hull of the $2n$ points that is neither $p$ nor $q$, and the same argument shows that there is a balanced line through it. Thus, there are at least two balanced lines.