, 2022

Proof: We first prove the claim: if the screen eventually turns into one color, say blue, then there must be one square which is constantly blue.
Suppose not, namely, suppose that eventually all squares are blue but every square has changed color at some time. We construct an oriented graph $G$ as follows: the vertices are all squares, and we link an oriented arrow $A \to B$ if $A$ and $B$ are adjacent and there is some time $t_0$ such that square $A$ is not blue at the time $t_0 - 1$ and is constantly blue when $t \ge t_0$, and square $B$ is blue at the time $t_0 - 1$. In other words, the square $A$ is changed to blue last time because of square $B$. Since we assume that every square on the screen has changed color, every vertex of the graph has out-degree $\ge 1$. So $G$ must contain an oriented loop $A_1 \to A_2 \to \dots \to A_k \to A_1$. Assume that square $A_i$ turns blue for the last time at the time $t_i$, i.e. $A_i$ is not blue at the time $t_i - 1$, and is constantly blue when $t \ge t_i$. Without loss of generality, $t_1$ is the largest among $t_1, \dots, t_k$. From the arrow $A_1 \to A_2$, the square $A_i$ is yellow at time $t_1 - 1$; From the arrow $A_k \to A_1$, the square $A_1$ is blue at the time $t_k - 1$. We must have $t_k < t_1$. But a blue square can only turn red first before turning yellow. So there exists $t_k < t' < t_1$, such that $A_1$ is red at the time $t' - 1$. But $t' - 1 \ge t_k$, $A_k$ is blue at the time $t' - 1$, so $A_1$ will force $A_k$ to turn red at the time $t'$. But this contradicts the definition of $t_k$, namely, $A_k$ is constantly blue from the time $t_k$ onward. The claim is proved.

Back to the original problem, assume that all squares are turned blue eventually, we will show that the squares are all turned blue after $2n - 2$ seconds.
By the claim, there exists a square $A$ which is constantly blue. Define the distance between two squares by the sum of their horizontal and vertical distances; then the distance between farthest two squares is $2n - 2$. If $B$ and $A$ have distance 1, then $B$ can never be red, this means: if $B$ is initially blue, then it is always blue; if $B$ is yellow, then

it is turned blue after 1 second and continues to be blue after that. Now, for a point $B$ of distant $k$ to the point $A$, we may run an induction to prove that $B$ will become constantly blue after $k$ seconds. Suppose that this statement has been proved for $k-1$. If $B$ has distant $k$ to $A$, take a square $B'$ adjacent to $B$ that has distant $k-1$ to $A$. By induction, $B'$ is constant blue after $k-1$ seconds, then the same argument as above shows that $B$ is constantly blue after $k$ seconds. From this induction, we deduce that all squares are blue after $2n-2$ seconds.