A $8\times 8$ board is given. Seven out of $64$ unit squares are painted black. Suppose that there exists a positive $k$ such that no matter which squares are black, there exists a rectangle (with sides parallel to the sides of the board) with area $k$ containing no black squares. Find the maximum value of $k$. 

1. **Understanding the Problem:**
   We are given an $8 \times 8$ board with 7 out of the 64 unit squares painted black. We need to find the maximum area $k$ of a rectangle (with sides parallel to the sides of the board) that contains no black squares, regardless of the placement of the black squares.

2. **Initial Consideration:**
   We need to ensure that for any configuration of 7 black squares, there exists a rectangle of area $k$ that is free of black squares. 

3. **Achievability of $k=8$:**
   Consider the possibility of $k=8$. This would mean finding a rectangle of area 8 that is free of black squares. Since there are 8 rows and 8 columns, if we place 7 black squares, at least one row must be completely free of black squares (by the pigeonhole principle). This row can then form a rectangle of area 8 (1 row × 8 columns).

4. **Proof by Example:**
   To see that $k=8$ is achievable, consider the following configuration of black squares:
   ```
   O O O O O O O O
   O O X O O X O O
   O O O O O O O O
   O X O O O O X O
   O O O X O O O O
   O O O O O O O O
   O O X O O X O O
   O O O O O O O O
   ```
   In this configuration, the second, fourth, and seventh rows have black squares, but the first, third, fifth, sixth, and eighth rows are completely free of black squares. Therefore, we can always find a row that is free of black squares, allowing us to form a rectangle of area 8.

5. **Impossibility of $k>8$:**
   To show that $k>8$ is not possible, consider the worst-case scenario where black squares are placed in such a way that they maximize the disruption of potential rectangles. For instance, if black squares are placed in every row and every column, it would be impossible to find a rectangle larger than 8 that is free of black squares. 

   For example, if black squares are placed as follows:
   ```
   X O O O O O O O
   O X O O O O O O
   O O X O O O O O
   O O O X O O O O
   O O O O X O O O
   O O O O O X O O
   O O O O O O X O
   O O O O O O O X
   ```
   In this configuration, every row and every column contains at least one black square, making it impossible to find a rectangle larger than 8 that is free of black squares.

Conclusion:
Thus, the maximum value of $k$ such that there always exists a rectangle of area $k$ free of black squares, regardless of the placement of the black squares, is $k=8$.

The final answer is $\boxed{8}$.