Example 1 In a family photo album, there are 10 photos. Each photo has 3 men, the person standing on the left is the son of the person in the middle, and the person on the right is the middle person's brother. It is known that the 10 people in the middle of the 10 photos are all different. How many different people are there at least in these 10 photos?
Solution: As shown in Figure 1, there are 16 people, where the horizontal lines represent brothers, and the vertical and diagonal lines represent father-son relationships. The 10 photos taken by them are as follows:
$$
\begin{array}{l}
\quad\{3,1,2\},\{16, \\
2,1\},\{5,3,4\},\{15, \\
4,3\},\{7,5,6\},\{9, \\
6,5\},\{11,7,8\}, \\
\{12,8,7\},\{13,9, \\
10\},\{14,10,9\} .
\end{array}
$$

The 10 people in the middle are exactly $1,2, \cdots, 10$, and they are all different. Therefore, the minimum number of people required is no more than 16.

On the other hand, we use the "grouping method" to prove that it is impossible to take 10 photos that meet the requirements with only 15 different people.

If not, suppose there are 15 people who can meet the requirements. We group these 15 people by brothers, placing brothers in the same group and non-brothers in different groups, and those without brothers form a group by themselves.
(1) Since the 10 people in the middle of the 10 photos are all different, i.e., the 10 fathers in the middle are all different, the 10 sons on the left of the photos are in different groups, i.e., they belong to 10 different groups. In addition, the person with the highest seniority among the 15 people in the 10 photos must belong to another group, so the 15 people must be divided into at least 11 groups.
(2) We examine the following two types of groups:
(i) Each person in the group has appeared in the middle position of a photo;
(ii) At least one person in the group has never appeared in the middle position of a photo.

Since the person on the right of each photo is the brother of the person in the middle, the group of the person in the middle must have at least 2 people. Since there are only 10 people in the middle, there can be at most 5 groups of type (i).

Because there are 15 people in the 10 photos, and only 5 people have never appeared in the middle position, there can be at most 5 groups of type (ii).
Thus, the 15 people can be divided into at most 10 groups, which is a contradiction.
In conclusion, there must be at least 16 different people in the 10 photos.