We can easily check cases $n=3,4,5$. Suppose $n \geq6$. Define as $P(n)$ the largest number of pairwise disjoint monochromatic segments we can always have. We need to show that $P(n+3) \geq P(n)+1$ (essentially it is induction with step 3). $Case 1$: There do not exist three consecutive vertices $A_i,A_{i+1},A_{i+2}$ such that the edges that connect them have both colors. Then, the perimeter of the convex polygon is monochromatic, and the result is obvious. $Case 2$: There exist three consecutive vertices $A_i,A_{i+1},A_{i+2}$ such that the edges that connect them have both colors. Then, take them aside and colour the rest $n-3$. Then, we get at least $P(n-3)$ pairwise disjoint monochromatic segments of color $a$. Since the edges that connect the 3 selected vertices are disjoint from the others, we can select one which is of color $a$ and we get $P(n)\geq P(n-3)+1$,done.