C15 (16-4, Bulgaria) Divide an $8 \times 8$ chessboard into $p$ rectangles, such that the division does not cut through any cell (i.e., it must follow the grid lines of the chessboard), and satisfies the following two conditions:
（1）Each rectangle contains the same number of white and black cells;
（2）If the number of white cells in the $i$-th rectangle is $a_{i}$, then $a_{1}<a_{2}<\cdots<a_{p}$. Find the maximum value of $p$ among all possible divisions, and for this maximum value of $p$, list all possible sequences $a_{1}, a_{2}, \cdots, a_{p}$.
Since the chessboard has 32 white squares, we have
$$
a_{1}+a_{2}+\cdots+a_{p}=32 \text {. }
$$

From (2), $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots, a_{p} \geqslant p$, so
$$
\begin{array}{c}
1+2+\cdots+p \leqslant 32, \\
p^{2}+p \leqslant 64,
\end{array}
$$

thus $p \leqslant 7$. That is, the chessboard can be divided into at most 7 rectangles that meet the conditions.
When $p=7$, there are five possible combinations:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$a_{1}$ & $a_{2}$ & $a_{3}$ & $a_{4}$ & $a_{5}$ & $a_{6}$ & $a_{7}$ \\
\hline (1) 1 & 2 & 3 & 4 & 5 & 6 & 11 \\
\hline (2) 1 & 2 & 3 & 4 & 5 & 7 & 10 \\
\hline (3) 1 & 2 & 3 & 4 & 5 & 8 & 9 \\
\hline (4) 1 & 2 & 3 & 4 & 6 & 7 & 9 \\
\hline (5) 1 & 2 & 3 & 5 & 6 & 7 & 8 \\
\hline
\end{tabular}

The first combination is impossible to achieve on the chessboard, as it is not possible to divide the chessboard into a 22-square rectangle. All other cases are possible, and specific divisions are shown in Figure 101 (1) and (4) (the numbers in the figure represent the number of squares in each block).