Instead of considering a complete graph $K_n$ whose edges are red and blue, just consider some graph $G$ with $n$ vertices. Let $\bar G$ be the complement of $G$. $\bar G$ contains a clique of size $s$ if and only if $G$ contains an independent set of size $s$.
Now consider $G$ as a subgraph of $K_n$. Color the edges of $G$ in $K_n$ red, and color the other edges of the $K_n$, that is the edges of $\bar G$, blue.
Now $K_n$ contains a red $K_r$ if and only if $G$ contains a clique of size $r$, and $K_n$ contains a blue $K_b$ if $\bar G$ contains a clique of size $b$, which is true if and only if $G$ contains an independent set of size $b$.
The Ramsey theorem says that for given $r$ and $b$ there is an $n$ such that (what you said about $K_n$). A Ramsey graph of size $q$ is a counterexample to the Ramsey theorem, and when it exists, it shows that $n$, which must exist, must be larger than $q$.
Here is an example. Let's take $r=b=3$. The Ramsey theorem says that there is some $n$ such that if we color edges of $K_n$ in red and blue, there is either a red triangle or a blue triangle.
There are many Ramsey graphs. For example, consider $K_2$. It has neither a clique of size $r=3$ nor an independent set of size $b=3$ and therefore it is a ramsey graph for $r=3, b=3$ and shows that the $n$ of the previous paragraph must be bigger than 2.
Now consider $C_5$, the cycle on five vertices. It has neither a clique of size $r=3$ nor an independent set of size $b=3$ and therefore it is a ramsey graph for $r=3, b=3$ and shows that the $n$ given by the Ramsey theorem must be bigger than 5.
The Ramsey theorem claims that when $n$ is large enough, there is no Ramsey graph with $n$ or more vertices.