Every cell of $8\times8$ chessboard contains either $1$ or $-1$. It is known that there are at least four rows such that the sum of numbers inside the cells of those rows is positive. At most how many columns are there such that the sum of numbers inside the cells   of those columns is less than $-3$?

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 2$
1. **Understanding the Problem:**
   - We have an $8 \times 8$ chessboard where each cell contains either $1$ or $-1$.
   - There are at least four rows where the sum of the numbers in the cells is positive.
   - We need to determine the maximum number of columns where the sum of the numbers in the cells is less than $-3$.

2. **Sum of Rows and Columns:**
   - The sum of any row or column must be one of $\{-8, -6, -4, -2, 0, 2, 4, 6, 8\}$ because each cell is either $1$ or $-1$.
   - For a column to have a sum less than $-3$, it must have at least six $-1$s (since $-4$ is the smallest sum less than $-3$).

3. **Analyzing the Rows:**
   - We know that there are at least four rows with a positive sum. This means each of these rows has more $1$s than $-1$s.
   - Let’s assume these four rows have the minimum number of $-1$s to still have a positive sum. The minimum positive sum is $2$, which means each of these rows has at most three $-1$s (since $4 \times 1 + 4 \times (-1) = 0$ and we need a positive sum).

4. **Constructing the Chessboard:**
   - Let’s denote the four rows with positive sums as $R_1, R_2, R_3, R_4$.
   - Each of these rows has at most three $-1$s. Therefore, each of these rows has at least five $1$s.
   - The remaining four rows can have any combination of $1$s and $-1$s.

5. **Maximizing the Number of Columns with Sum Less than $-3$:**
   - To maximize the number of columns with a sum less than $-3$, we need to place as many $-1$s as possible in the columns.
   - Since $R_1, R_2, R_3, R_4$ each have at most three $-1$s, we can place these $-1$s in different columns to ensure that no column in these rows has more than three $-1$s.
   - The remaining four rows can be filled entirely with $-1$s to maximize the number of columns with a sum less than $-3$.

6. **Constructing an Example:**
   - Let’s construct an example where the first four rows have three $-1$s each, and the last four rows have all $-1$s:
     \[
     \begin{array}{cccccccc}
     -1 & -1 & -1 & 1 & 1 & 1 & 1 & 1 \\
     -1 & -1 & -1 & 1 & 1 & 1 & 1 & 1 \\
     -1 & -1 & -1 & 1 & 1 & 1 & 1 & 1 \\
     -1 & -1 & -1 & 1 & 1 & 1 & 1 & 1 \\
     -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
     -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
     -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
     -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
     \end{array}
     \]
   - In this configuration, the first three columns have a sum of $-8$ (since each column has eight $-1$s), and the remaining columns have a sum of $0$ or positive.

7. **Conclusion:**
   - We can see that we can have at most six columns where the sum of the numbers inside the cells is less than $-3$.

The final answer is $\boxed{6}$