Example: 510 people go to the bookstore to buy books, it is known that (1) each person bought 3 types of books; (2) any two people have at least one book in common. How many people at most bought the most purchased book? Explain your reasoning.

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Solution: In the right figure, there are 5 triangles formed by the center and two adjacent vertices of the regular pentagon, and there are also 5 triangles formed by 3 non-adjacent vertices of the regular pentagon. It is not difficult to see that any two of these 10 triangles share at least one common vertex. Write out the vertex number groups of these triangles and let the book numbers purchased by 10 people be these 10 vertex number groups of the triangles:
(123), (134), (145), (156), (162), (245), (356), (426), (523), (634).

Obviously, each book is purchased by someone. Therefore, the required minimum value does not exceed 5.

Assume the required minimum value is 4, and 10 people buy a total of $n$ types of books, with the $i$-th type of book purchased by $m_{i}$ people, so $m_{i} \leqslant 4$ and $m_{1}+$ $m_{2}+\cdots+m_{n}=30$. When two people buy the same book, it is called a "book pair". According to the given information, there are at least $\mathrm{C}_{10}^{2}=45$ book pairs. On the other hand, the number of book pairs formed by the $i$-th type of book is $\mathrm{C}_{m_{i}}^{2}$, totaling $\mathrm{C}_{m_{1}}^{2}$ $+\mathrm{C}_{m_{2}}^{2}+\cdots+\mathrm{C}_{m_{n}}^{2}$ book pairs. Therefore, we have
$$
\mathrm{C}_{m_{1}}^{2}+\mathrm{C}_{m_{2}}^{2}+\cdots+\mathrm{C}_{m_{n}}^{2} \geqslant 45
$$

Since $\mathrm{C}_{4}^{2}=6, \mathrm{C}_{3}^{2}=3, \mathrm{C}_{2}^{2}=1$, we also have
$$
\mathrm{C}_{m_{1}}^{2}+\mathrm{C}_{m_{2}}^{2}+\cdots+\mathrm{C}_{m_{n}}^{2} \leqslant 7 \mathrm{C}_{4}^{2}+\mathrm{C}_{2}^{2}=43
$$

Since (1) and (2) are contradictory, the required minimum value is 5.