4. (6) In each cell of a $100 \times 100$ table, a number from the interval $[-1,1]$ is written. The sum of all the numbers is 0, the sum of all the numbers in each row in absolute value is at least $c$, and the sum of all the numbers in each column in absolute value is at least $c$. Find the largest possible value of $c$.
4. Answer: 50. Example. Divide the square into four squares with side 50, place zeros in two opposite squares, ones in one, and minus ones in the other.

Let's prove that it cannot be more. Suppose that in each row and each column the sum is greater than 50 in absolute value. Let there be $50+x$ columns with a positive sum and $50-x-$ with a negative sum ( $-49 \leqslant x \leqslant 49$ ). Note that then the sum in the "positive" columns is greater than $50(50+x)$, and in the "negative" ones - less than $-50(50-x)$. On the other hand, these two sums are opposite numbers, so in any case we get that the sum in the positive columns is greater than $50(50+|x|)$. Similarly, if there are $50+y$ positive rows, then the sum of the numbers in them is greater than $50(50+|y|)$. Denote by $A, B, C$ the sums of the numbers in the cells at the intersection of positive columns and positive rows; positive columns and negative rows; negative columns and negative rows. We have $A+$ $B>50(50+|x|), B+C>50(50+|y|),(|x|-25)(|y|-25)>25^{2}$, which is obviously impossible. Contradiction.