4
RBRB

2

We use a string consisting of `R` and `B`, corresponding to red and blue, to represent the sequence of colors of the chips of a pile from top to bottom. Particularly, the sequence of colors of the chips of a $0$-chip pile is an empty string $\varepsilon$. Also, the state where the sequences of colors of chips of the piles on square $1, 2, 3, 4$ are $S_1, S_2, S_3, S_4$, respectively, is called as state $(S_1, S_2, S_3, S_4)$. The following process yields `RRBB` as the sequence of colors of chips of the pile on square $4$. - First, on each square, place a chip of the same color as the square, resulting in state $($ `R` $,$ `B` $,$ `R` $,$ `B` $)$. - Place the pile of chips on square $1$ on top of the pile of chips on square $2$, resulting in state $(\varepsilon,$ `RB` $,$ `R` $,$ `B` $)$. - Place the pile of chips on square $2$ on top of the pile of chips on square $4$, resulting in state $(\varepsilon, \varepsilon,$ `R` $,$ `RBB` $)$. - Place the pile of chips on square $3$ on top of the pile of chips on square $4$, resulting in state $(\varepsilon, \varepsilon, \varepsilon,$ `RRBB` $)$. The following process yields `RBRB` as the sequence of colors of chips of the pile on square $4$. - First, on each square, place a chip of the same color as the square, resulting in state $($ `R` $,$ `B` $,$ `R` $,$ `B` $)$. - Place the pile of chips on square $1$ on top of the pile of chips on square $4$, resulting in state $(\varepsilon,$ `B` $,$ `R` $,$ `RB` $)$. - Place the pile of chips on square $2$ on top of the pile of chips on square $4$, resulting in state $(\varepsilon, \varepsilon,$ `R` $,$ `BRB` $)$. - Place the pile of chips on square $3$ on top of the pile of chips on square $4$, resulting in state $(\varepsilon, \varepsilon, \varepsilon,$ `RBRB` $)$. The above two, `RRBB` and `RBRB`, are the only possible sequences of colors of chips of the pile on square $4$ from top to bottom after the process in the problem.