The answer is $n=1280$. Assign red and green points with $1$ and $0$. If the points are monochromatic, then $n=1$ suffices. Now assume otherwise. Note that in each step, $\cdots 010 \cdots$ changes to $\cdots 101 \cdots$; $\cdots 0110 \cdots$ changes to $\cdots 1001 \cdots$; $\cdots 0 \underbrace{11 \cdots 11}_{n} 0 \cdots$ changes to $\cdots 10 \underbrace{1 \cdots 1}_{n-2} 01 \cdots$ for $n \ge 2$. In the third case $\ell := \text{length of longest consecutive monochromatic points}$ decreases $2$. Initially $\ell_1 \le 2560$, so $\ell_{1280} \le 2$, hence only the first and second cases happen after $F_{1280}$ and it become $2-\text{periodic}$. The result is sharp if the initial coloring is $1$ red and $2560$ greens.