Solving the system $a+ \dfrac{1} {b} = x$, $b+ \dfrac {1} {a} = y$, for $x,y>1$, leads to $a = \dfrac {xy \pm \sqrt{xy(xy-4)}} {2y}$, $b = \dfrac {xy \pm \sqrt{xy(xy-4)}} {2x}$, requiring $xy \geq 4$ for real solution(s). Thus we need show there exist points $x,y > 1$ differently colored, with $xy\geq 4$. The point $2$ has some color; if any other point $c>2$ has a different color, we may take $(x,y) = (2,c)$. If not, it means all other points $c>2$ have same color with $2$. But there will exist some point $1<x<2$ of a different color than $2$, and then, since $\dfrac {4} {x} > 2$, we can take $y=\dfrac {4} {x}$.

Thanks, mavropnevma. As usual, a nice and clearly-explained solution. How did you possibly solve and TeX up such a thing only five minutes after the OP? This contest had a different but easier and similar enough problem. Rather than wasting a new topic, I'm posting it here so that I have a link to get the entire contest posted in the "Resources" section. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Another Problem: Each real number greater than $1$ is colored red or blue with both colors being used. Prove that there exist real numbers $a$ and $b$ such that the numbers $a+b$ and $ab$ are of different colors.

Solving the system $a + b = x$, $ab = y$, for $x,y>1$, leads to $a = \dfrac {x \pm \sqrt{x^2-4y}} {2}$, $b = \dfrac {x \mp \sqrt{x^2 - 4y}} {2}$, requiring $x^2 \geq 4y$ for real solution(s). Thus we need show there exist points $x,y > 1$ differently colored, with $x^2\geq 4y$. The point $4$ has some color; if any other point $c>4$ has a different color, we may take $(x,y) = (c,4)$. If not, it means all other points $c>4$ have same color with $4$. But there will exist some point $1<y<4$ of a different color than $4$, and then we can take $x=4$.