Seems to be not answered for long enough, so let me post a solution. If $\mathcal{A}$, $\mathcal{B}$ denote the arrangements enumerated by $A$ and $B$, respectively, we construct a bijection between $\mathcal{A}\setminus \mathcal{B}$ and $\mathcal{B}\setminus \mathcal{A}$. Let the red row contain $r_1\geqslant r_2\geqslant \ldots$ balls, the blue row $b_1\geqslant b_2\geqslant \ldots$ balls. Then $\mathcal{A}$ is the set of arrangements for which $b_1>r_1$, and $\mathcal{B}$ is the set of arrangements for which $b_i=i$ for certain $i$. Take an arrangement in $\mathcal{A}\setminus \mathcal{B}$: then, $b_1>r_1$ and $b_i\ne i$ for all $i$. The last condition means that for some $j$, we have $b_j>j$, $b_{j+1}<j+1$. Replace $b$'s to $b_2-1\geqslant \ldots b_j-1\geqslant j\geqslant b_{j+1}\ldots$ and $r$'s to $b_1-1\geqslant r_1\geqslant r_2\geqslant \ldots$. The new arrangement belongs to $\mathcal{B}$ (since the $j$-th blue basket contains $j$ balls) and does not belong to $\mathcal{A}$ (since the first red basket contains $b_1-1\geqslant b_2-1$ balls). The constructed map from $\mathcal{A}\setminus \mathcal{B}$ to $\mathcal{B}\setminus \mathcal{A}$ is indeed a bijection, the inverse maps takes $j$ for which $b_j=j$, removes this basket, increases by 1 the number of balls in first $j-1$ blue baskets and in the largest red basket, and moves this largest red basket to the blue row.