15. There is a 6-row $n$-column matrix composed of 0s and 1s, where each row contains exactly 5 ones, and the number of columns in which any two rows both have a 1 is at most 2. Find the minimum value of $n$.
15. First, the total number of 1s in the matrix is $5 \times 6=$ 30.
Let the number of 1s in the $k$-th column be $l_{k}$. Then
$$
\sum_{k=1}^{n} l_{k}=30 \text {. }
$$

For any $1 \leqslant i<j \leqslant 6$ and $1 \leqslant k \leqslant n$, consider the triplet $\{i, j, k\}$: such that the $i$-th row and the $j$-th row in the $k$-th column are both 1. The number of such triplets $\{i, j, k\}$ in the $k$-th column is $\mathrm{C}_{l_{k}}^{2}$, so the total number of such triplets $\{i, j, k\}$ is $\sum_{k=1}^{n} \mathrm{C}_{l_{k}}^{2}$.

Second, we calculate the total number of such triplets $\{i, j, k\}$ using another method: for any $1 \leqslant i<j \leqslant 6$, let $m_{i, j}$ be the number of columns where the $i$-th row and the $j$-th row are both 1, then
$$
\sum_{1 \leqslant i<j<6} m_{i, j}=\sum_{k=1}^{n} \mathrm{C}_{l_{k}}^{2} \text {. }
$$

Thus, by the condition $m_{i, j} \leqslant 2$, we have
$$
\begin{array}{l}
30=\frac{6(6-1)}{2} \times 2 \geqslant \sum_{1 \leqslant i<j \leqslant 6} m_{i, j} \\
=\sum_{k=1}^{n} \mathrm{C}_{l_{k}}^{2}=\frac{1}{2}\left(\sum_{k=1}^{n} l_{k}^{2}-\sum_{k=1}^{n} l_{k}\right) \\
\geqslant \frac{1}{2}\left[\frac{1}{n}\left(\sum_{k=1}^{n} l_{k}\right)^{2}-\sum_{k=1}^{n} l_{k}\right] \\
=\frac{1}{2 n} \times 30^{2}-\frac{1}{2} \times 30,
\end{array}
$$

which implies $n \geqslant 10$.
For $n=10$, Figure 2 is a matrix that meets the requirements:
$$
\begin{array}{llllllllll}
1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0
\end{array}
$$

Figure 2
Therefore, the minimum value of $n$ is 10.
(Fu Longrang provided)