Let $A_i$ be the segment with exactly $3n$ points from $S$, the leftmost being the $i$-th point from left to right of $S$, $i = 1, 2, \dots, 3n + 1$. Define $f(i)$ as the number of blue points in $A_i$. We have to prove that $f(i) = 2n$ for some $i$.

Notice that $|f(i + 1) - f(i)| \le 1$ since $A_i$ and $A_{i+1}$ have $3n - 1$ common points and $f(1) + f(3n+1) = 4n$ because the disjoint segments $A_1$ and $A_{3n+1}$ cover $S$.

If $f(1) = 2n$ we are done. So suppose without loss of generality that $f(1) < 2n$. So $f(3n + 1) > 2n$ and, since $f(i)$ increases or decreases at most 1, there must exist a number $k$ such that $f(k) = 2n$.