37th Hellenic Mathematical Olympiad 2020

(a) Each circle can be colored with two different colors independently of the coloring of the other circles. Therefore, according to the multiplicative principle, the different colorings are $$\underset{99}{\underbrace{2\cdot 2\cdot 2\cdots 2}}=2^{99}.$$

(b) We consider the coloring $X:123\dots 50\underset{x}{\underbrace{51\cdots 9899}}$, where $x$ is the number of red circles among $1$ to $50$, and $y$ is the number of red circles among $51$ to $99$. The coloring $X$ is «good» if $x>y$, whereas the coloring $X$ is no good if $x\le y$.

Let $A$ be the set of «good» colorings and $B$ the set of “no good” colorings. We will prove that each element of the set $A$ corresponds to an element of $B$ and vice versa.

Indeed, if $X:123\dots 50\underset{x}{\underbrace{51\cdots 9899}}$ is in $A$, then $x>y$. By changing the color of each circle, we find the coloring $$Y:\underset{50-x}{\underbrace{123\dots 50}}\underset{49-y}{\underbrace{51\cdots 9899}},$$ which belongs to $B$, because $$x>y\Rightarrow -x<-y\Rightarrow 49-x<49-y\Rightarrow 50-x\le 49-y.$$

Conversely, if $X:123\dots 50\underset{x}{\underbrace{51\cdots 9899}}$ is in $B$, then $x\le y$. By changing the color of each circle, we find the coloring $$Y:\underset{50-x}{\underbrace{123\dots 50}}\underset{49-y}{\underbrace{51\cdots 9899}},$$ which belongs to the set $A$ because $$x\le y\Rightarrow -x\ge -y\Rightarrow 49-x\ge 49-y\Rightarrow 50-x>49-y.$$

Therefore, between the sets $A$ and $B$ there exists a $1$-$1$ correspondence and so the two sets have the same number of elements, that is, each of them has $$\frac{2^{99}}{2}=2^{98}\text{ elements.}$$