$5 \cdot 579$ people go to the bookstore to buy books, and each person buys 4 kinds of books. Every two people have exactly two kinds of books in common. How many kinds of books did the 9 people buy in total? Explain your reasoning.
[Solution 1] Let the 4 books bought by A and B be (1,2,3,4) and (1,2,5,6) respectively.
(1) If someone did not buy books 1 and 2, then the 4 books they bought can only be (3,4,5,6). Therefore, there can be at most 1 such person, and the remaining 6 people must have bought at least one of books 1 and 2. Suppose C, D, and E all bought book 1. If one of these 3 people bought book 2, then at least 3 people bought both books 1 and 2; if none of these 3 people bought book 2, then since each of them bought 4 books that are exactly two the same as A's, all 3 must buy one of 3 and 4. Suppose two of them bought book 3, then 3 people bought both books 1 and 3.
(2) From the above proof, we can assume that A, B, and C bought the books (1,2,3,4), (1,2,5,6), and (1,2,7,8) respectively. It is easy to see that the remaining 6 people all bought at least one of books 1 and 2. Suppose D, E, and F all bought book 1. If one of these 3 people bought book 2, then at least 4 people bought both books 1 and 2. If none of these 3 people bought book 2, then each of them must buy one of 3 and 4, one of 5 and 6, and one of 7 and 8. If D, E, and F bought the same one from {3,4,5,6,7,8}, then there would be at least 4 people who bought the same two books. Otherwise, the 4 books bought by D, E, and F can only be (1,3,5,7), (1,3,6,8), (1,4,5,8), and (1,4,6,7) in 3 groups. Suppose they are the first 3 groups. Clearly, at least two of G, H, and I bought book 2. Since two of them bought (1,3,5,7) and (1,3,6,8), the two who bought book 2 cannot buy book 4, so they can only buy books 2 and 3. Compared with (1,4,6,7), they can only buy (2,3,6,7). But two people cannot buy the same, leading to a contradiction. This indicates that at least 4 people bought the same two books.
(3) Suppose A, B, C, and D bought the books (1,2,3,4), (1,2,5,6), (1,2,7,8), and (1,2,9,10) respectively. If one of the remaining 5 people did not buy both books 1 and 2, suppose E bought 1 but not 2, then he must take one from each of the 4 pairs (3,4), (5,6), (7,8), (9,10), leading to E buying 5 books, which is a contradiction. This means that each of the 9 people bought both books 1 and 2. Since each pair of people bought 4 books that are exactly two the same, the 9 people together bought 20 books.
[Solution 2] According to the given, A and each of the other 8 people bought exactly 2 books in common, totaling 16 books. But A bought 4 books. By the pigeonhole principle and further analysis, either 5 of the other 8 people, along with A, bought the same book, or the other 8 people formed 4 groups of four, each group of 5 people including A bought the same book. For the other 8 people, it is also this situation. If there is no case where 6 people bought the same book, then the number of people who bought each book is exactly 5, and the total number of books bought should be a multiple of 5. But the total number of books bought by 9 people is 36, which is a contradiction. Therefore, there must be 6 people who bought the same book.

Suppose A, B, C, D, E, and F all bought book α. If there are two people G and H who did not buy α, then because G and H each have 12 books in common with the first 6 people, either 4 of the first 6 people, along with G or H, bought the same book, or the 6 people are divided into 4 groups of three, each group of 3 people with G (or H) bought the same book.
(1) If it is the former, suppose A, B, C, D, and G all bought book β, then besides β, G and A, B, C, D each have one book in common. Since G bought 3 books besides β, there must be two of A, B, C, D who bought the same book γ with G. Then these two people bought 3 books in common: α, β, γ, which is a contradiction.
(2) If it is the latter, i.e., the 6 people are divided into 4 groups of three (each person is in exactly two groups), each group of 3 people with G (or H) bought the same book. Suppose the 4 books bought by G and H are (β, γ, δ, ε) and (β, γ, θ, φ) respectively. Then the union of the two groups of three people who bought δ and ε is the same as the union of the two groups of three people who bought θ and φ. By the pigeonhole principle, among the 3 people who bought θ, there are two who also bought δ or ε. This leads to two people buying 3 books in common, which is a contradiction. This indicates that at least 8 people bought book α.

Suppose A, B, C, D, E, F, G, and H bought book α. Besides α, the 8 people, the last 7 of whom each have one book in common with A, are one of A's other 3 books. By the pigeonhole principle, there must be 3 people, plus A, 4 people bought the same book β. Then 4 people bought books α and β. Therefore, everyone bought books α and β. So, the 9 people together bought 20 books.