9. 4. In a family photo album, there are 10 photos. Each photo has 3 people, with the man standing in the middle, his son standing on his left, and his brother on his right. If it is known that the 10 men standing in the middle are all different, then what is the minimum number of different people in these photos?
9. 4. 16.

We refer to the 10 men standing in the middle of the photo as the main characters. We then classify all the men in the photo into levels: men who do not have a father in any photo are classified as level 0; when $k=0,1,2$, $\cdots$, men who have a father classified as level $k$ in some photo are classified as level $k+1$. Let $r_{k}$ represent the number of main characters at level $k$, and $t_{k}$ represent the number of other men at level $k$. The number of fathers of level $k+1$ men does not exceed $\frac{1}{2} r_{k-1} + t_{k+1}$, because each main character has brothers. At the same time, the number of fathers of level $k-1$ men is no less than $r_{k}$, because each main character has a son. Therefore,
$$
r_{k} \leqslant \frac{1}{2} r_{k+1} + t_{k+1}, \quad k=0,1,2, \cdots.
$$

Additionally, $1 \leqslant \frac{1}{2} r_{0} + t_{0}$. Adding all the inequalities together, we get
$$
\frac{1}{2}\left(r_{0} + r_{1} + \cdots\right) + 1 \leqslant t_{0} + t_{1} + \cdots,
$$

Thus, $\left(r_{0} + r_{1} + \cdots\right) + \left(t_{0} + t_{1} + \cdots\right)$
$$
\geqslant \frac{3}{2}\left(r_{0} + r_{1} + \cdots\right) + 1 = \frac{3}{2} \cdot 10 + 1 = 16.
$$

Therefore, there are at least 16 men in the photo.
There are 10 main characters (numbered 1 to 10) and 16 men in the photo, as shown in the diagram. Horizontal lines connect brothers, while other lines (from top to bottom) connect fathers and sons.

The 10 photos are: $(3,1,2)$; $(5,2,1)$; $(7,3,4)$; $(9,4,3)$; $(11,5,6)$; $(12,6,5)$; $(13,7,8)$; $(14,8,7)$; $(15,9,10)$; $(16,10,9)$.