For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.

How many people at most could have bought the book that was purchased by the fewest people?
(1993, China Mathematical Olympiad)
Solution: Let's assume that person A bought three books, and since A has at least one book in common with each of the other 9 people, among A's three books, the book purchased by the most people is bought by no less than 4 people.

If the book purchased by the most people is bought by 4 people, then all three books bought by A are each purchased by 4 people, and each book bought by the other 9 people is also purchased by 4 people. Therefore, the total number of books bought by the 10 people is a multiple of 4. That is, $4 \mid 30$, which is a contradiction. Thus, the book purchased by the most people is bought by at least 5 people.
Consider the following purchasing method:
$$
\begin{array}{l}
\left\{B_{1}, B_{2}, B_{3}\right\},\left\{B_{1}, B_{2}, B_{4}\right\},\left\{B_{2}, B_{3}, B_{5}\right\}, \\
\left\{B_{1}, B_{3}, B_{6}\right\},\left\{B_{1}, B_{4}, B_{5}\right\},\left\{B_{2}, B_{4}, B_{6}\right\}, \\
\left\{B_{3}, B_{4}, B_{5}\right\},\left\{B_{1}, B_{5}, B_{6}\right\},\left\{B_{2}, B_{5}, B_{6}\right\}, \\
\left\{B_{3}, B_{4}, B_{6}\right\},
\end{array}
$$

It can be seen that the book purchased by the most people is bought by at least 5 people.