7. Four people went to the bookstore to buy books. Each person bought 4 different books, and every two people have exactly 2 books in common. Therefore, these 4 people bought at least $\qquad$ kinds of books.
Answer: 7.
Think from a simple situation, if there are only two people, then to meet the requirements of the question, there must be 6 different books. We can number these 6 books as $1, 2, 3, 4, 5, 6$. Suppose A buys $1, 2, 3, 4$, and B buys $1, 2, 5, 6$. At this point, a third person arrives, we call him C. To meet the requirements, C can choose not to buy other books, and buying the numbers $3, 4, 5, 6$ also meets the requirements. At this point, it is important to note that when there are three people, and only 6 books are bought, the numbers C buys are unique after A and B's purchases are determined. C cannot buy a book that both A and B have bought; if he buys a book that both A and B have bought, he must then buy another book that he "uniquely" owns! The following table makes what I just said clearer:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline A & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & & \\
\hline B & $\checkmark$ & $\checkmark$ & & & $\checkmark$ & $\checkmark$ \\
\hline C & & & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ \\
\hline
\end{tabular}

After understanding the situation with three people, let's look at four people. Now D arrives, thinking from C's perspective, he must have at least one "unique" book. Therefore, with four people, there must be at least 7 books. The construction can be given as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline A & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & & & \\
\hline B & $\checkmark$ & $\checkmark$ & & & $\checkmark$ & $\checkmark$ & \\
\hline C & & & $\checkmark$ & $\checkmark$ & $\checkmark$ & $\checkmark$ & \\
\hline D & $\checkmark$ & & $\checkmark$ & & $\checkmark$ & & $\checkmark$ \\
\hline
\end{tabular}

That is, A buys books numbered $(1,2,3,4)$, then B buys books numbered $(1,2,5,6)$, to minimize the number of books, C buys books numbered $(3,4,5,6)$, at this point D can buy $(1,3,5,7)$.