4. Consider an $n \times n$ matrix, where each element is a real number with absolute value not exceeding 1, and the sum of all elements is 0. Let $n$ be a positive even number. Find the minimum value of $c$ such that for every such matrix, there exists a row or a column whose sum of elements has an absolute value not exceeding $c$.


4. The minimum value of $c$ is $\frac{n}{2}$.

Define the matrix $\left(a_{i j}\right)_{1 \leqslant i, j \leqslant n}$ as follows:
$$
a_{i j}=\left\{\begin{array}{ll}
1, & i, j \leqslant \frac{n}{2} ; \\
-1, & i, j>\frac{n}{2} ; \\
0, & \text { otherwise. }
\end{array}\right.
$$

Then the sum of all elements is 0, and the sum of elements in each row and each column is $\pm \frac{n}{2}$. Therefore, $c$ is not less than $\frac{n}{2}$.

Next, we prove that for any $n \times n$ matrix satisfying the conditions, there exists a row or a column whose sum of elements lies in the interval $\left[-\frac{n}{2}, \frac{n}{2}\right]$. Assume the conclusion is false, i.e., the sum of elements in each row and each column is either greater than $\frac{n}{2}$ or less than $-\frac{n}{2}$. Without loss of generality, assume that at least $\frac{n}{2}$ rows have a positive sum.

Select $\frac{n}{2}$ rows with a positive sum. These $\frac{n}{2}$ rows form a $\frac{n}{2} \times n$ submatrix $M$, and the sum of all elements in $M$, $S_{M}$, is greater than $\frac{n^{2}}{4}$. For a column in $M$ with a positive sum, since each of the $\frac{n}{2}$ elements in the column is at most 1, their sum is at most $\frac{n}{2}$. To have $S_{M}>\frac{n^{2}}{4}$, at least $\frac{n}{2}$ columns in $M$ must have a positive sum. Therefore, we can select $\frac{n}{2}$ columns in $M$ with a positive sum, denoted as $c_{1}, c_{2}, \cdots, c_{\frac{n}{2}}$.

Let $N$ be the submatrix of the original matrix formed by the columns containing $c_{1}, c_{2}, \cdots, c_{\frac{n}{2}}$. Since the elements in the original matrix not in $M$ are at least -1, the sum of elements in each column of $N$ is greater than $-\frac{n}{2}$, and thus greater than $\frac{n}{2}$. Therefore, the sum of all elements in $N$, $S_{N}$, is greater than $\frac{n^{2}}{4}$. This means the sum of all elements in $M \cup N$ is also greater than $\frac{n^{2}}{4}$. In fact, since $S_{M}>\frac{n^{2}}{4}$, $S_{N}>\frac{n^{2}}{4}$, and the sum of all elements in $M \cap N$ is at most $\frac{n^{2}}{4}$ (since $M \cap N$ is a $\frac{n}{2} \times \frac{n}{2}$ submatrix with each element at most 1), we have:
$$
S_{M \cup N}=S_{M}+S_{N}-S_{M \cap N}>\frac{n^{2}}{4}.
$$

Now consider the elements that are neither in $M$ nor in $N$. There are $\frac{n^{2}}{4}$ such elements, each at least -1, so their sum is at least $-\frac{n^{2}}{4}$. Therefore, the sum of all elements in the original matrix is positive, which is a contradiction.