Consider decompositions of an $8\times 8$ chessboard into $p$ non-overlapping rectangles subject to the following conditions:
(i) Each rectangle has as many white squares as black squares.
(ii) If $a_i$ is the number of white squares in the $i$-th rectangle, then $a_1<a_2<\ldots <a_p$. 
Find the maximum value of $p$ for which such a decomposition is possible. For this value of $p$, determine all possible sequences $a_1,a_2,\ldots ,a_p$.
1. **Understanding the Problem:**
   We need to decompose an $8 \times 8$ chessboard into $p$ non-overlapping rectangles such that:
   - Each rectangle has an equal number of white and black squares.
   - The number of white squares in each rectangle, denoted as $a_i$, satisfies $a_1 < a_2 < \ldots < a_p$.

2. **Initial Constraints:**
   - The total number of squares on the chessboard is $64$.
   - Since each rectangle must have an equal number of white and black squares, each rectangle must cover an even number of squares.
   - Therefore, the total number of white squares in all rectangles combined is $32$.

3. **Bounding \( p \):**
   - We need to find the maximum value of $p$ such that the sum of the first $p$ terms of an increasing sequence of positive integers equals $32$.
   - The sum of the first $p$ positive integers is given by $\frac{p(p+1)}{2}$.
   - We need $\frac{p(p+1)}{2} \leq 32$.

4. **Solving for \( p \):**
   \[
   \frac{p(p+1)}{2} \leq 32
   \]
   \[
   p(p+1) \leq 64
   \]
   - Solving the quadratic inequality:
   \[
   p^2 + p - 64 \leq 0
   \]
   - Using the quadratic formula $p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 1$, $b = 1$, and $c = -64$:
   \[
   p = \frac{-1 \pm \sqrt{1 + 256}}{2}
   \]
   \[
   p = \frac{-1 \pm 17}{2}
   \]
   - This gives $p = 8$ or $p = -9$. Since $p$ must be positive, we have $p = 8$.

5. **Verification:**
   - If $p = 8$, then $\frac{8 \cdot 9}{2} = 36$, which exceeds $32$. Therefore, $p = 8$ is not possible.
   - Thus, the maximum value of $p$ is $7$.

6. **Finding Sequences:**
   - We need sequences $(a_1, a_2, \ldots, a_7)$ such that $a_1 + a_2 + \ldots + a_7 = 32$ and $a_1 < a_2 < \ldots < a_7$.
   - Possible sequences are:
     \[
     (1, 2, 3, 4, 5, 8, 9), \quad (1, 2, 3, 4, 5, 7, 10), \quad (1, 2, 3, 4, 6, 7, 9), \quad (1, 2, 3, 5, 6, 7, 8)
     \]

7. **Verification of Sequences:**
   - Each sequence sums to $32$ and satisfies the condition $a_1 < a_2 < \ldots < a_7$.
   - The sequence $(1, 2, 3, 4, 5, 6, 11)$ is not possible because it would require a rectangle of area $22$ with side lengths less than or equal to $8$, which is impossible.

The final answer is \( \boxed{ p = 7 } \) and the possible sequences are: