8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there exists a row where the cells intersecting with these two columns are both filled with 1. Let $x=\max _{1 \leqslant i \leqslant m}\left\{x_{i}\right\}$. Then $x_{\text {min }}=$ $\qquad$ .
8. 5 .

Construct the application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."

Let the $i(i=1,2, \cdots, m)$-th kind of book be bought by $x_{i}$ people, and $x=\max _{1 \leqslant i \leqslant m} x_{i}$. Then
$$
\sum_{i=1}^{m} x_{i}=30 \text {, and } x_{i} \leqslant x(i=1,2, \cdots, m) \text {. }
$$

Considering the total number of "same book pairs" between any two people, we get the inequality
$$
\begin{array}{l}
\sum_{i=1}^{m} \mathrm{C}_{x_{i}}^{2} \geqslant \mathrm{C}_{10}^{2} \\
\Rightarrow \sum_{i=1}^{m} \frac{x_{i}\left(x_{i}-1\right)}{2} \geqslant 45 .
\end{array}
$$

Thus, from conclusion (1) and equation (2), we have
$$
\begin{array}{l}
45 \leqslant \frac{1}{2}(x-1) \sum_{i=1}^{m} x_{i}=15(x-1) \\
\Rightarrow x \geqslant 4 .
\end{array}
$$

If $x=4$, then all $x_{i}=4(i=1,2, \cdots, m)$, which contradicts $\sum_{i=1}^{m} x_{i}=30$. Therefore, $x \geqslant 5$.

Below, we construct a scenario where $x=5$, and the book purchased by the most people is bought by exactly 5 people.
Let the books be $B_{i}$, and the 10 people's book purchasing situation is as follows:
$$
\begin{array}{l}
B_{1} B_{2} B_{3}, B_{3} B_{4} B_{5}, B_{1} B_{5} B_{6}, B_{1} B_{3} B_{5}, \\
B_{1} B_{2} B_{4}, B_{1} B_{4} B_{6}, B_{2} B_{4} B_{5}, B_{2} B_{5} B_{6}, \\
B_{2} B_{3} B_{6}, B_{3} B_{4} B_{6} .
\end{array}
$$

Therefore, the book purchased by the most people is bought by at least 5 people.