Croatia_2018

Assume the contrary, i.e. that there is no square bordered by three or four blue segments. Then each square is bordered by at least two red segments.

Now we count the pairs $(P,r)$, where $P$ is a unit square, and $r$ is a red segment adjacent to $P$. We will count them in two different ways.

Since every square is bordered by at least two red segments, we have at least $2n^2$ such pairs. On the other hand, each segment on the edge of the board has only one adjacent square, while the interior segments have two adjacent squares. This shows that we can express the number of pairs $(P,r)$ as $V+2U$, where $V$ is the number of red segments on the edge of the board, and $U$ the number of interior red segments. Thus $2n^2\le 2U+V$.

Furthermore, since there are no more than $n^2$ red segments, we also have $U+V\le n^2$. Adding these two inequalities we get $V=0$ and $U=n^2$.

This shows that the inequalities we derived are in fact equalities, so that each square is bordered by exactly two red segments and all the red segments are interior (not on the edge of the board).

Now we checker the board, black-white. Notice that each white square contains exactly two red segments, and that each red segment belongs to a uniquely determined white square. This means that there are $2B$ red segments, where $B$ is the number of white squares. However, this leads to a contradiction because we know that there are exactly $n^2$ red segments, and $n^2$ is odd.

This shows that our assumption was wrong, so there must be some unit square with at least three blue adjacent segments.