$\mathrm{B}-1$. Let $S$ be a set of $n$ distinct real numbers, and $A_{s}$ be the set of all distinct averages of pairs of elements from $S$. For a given $n \geqslant 2$, what is the minimum number of elements that $A_{s}$ can have?
$A_{s}$ contains at least $2 n-3$ elements.
Let $x_{1}<x_{2}<\cdots<x_{n}$ denote the elements of $S$, then
$$
\begin{array}{l}
\frac{x_{1}+x_{2}}{2}<\frac{x_{1}+x_{3}}{2}<\cdots<\frac{x_{1}+x_{n}}{2}<\frac{x_{2}+x_{n}}{2}< \\
\frac{x_{3}+x_{n}}{2}<\cdots<\frac{x_{n-1}+x_{n}}{2}
\end{array}
$$

represents $(n-1)+(n-2)=(2 n-3)$ distinct elements in $A_{s}$, hence, $A_{s}$ contains at least $2 n-3$ distinct elements.

If we take $S=\{1,2, \cdots, n\}$, then the elements in $A_{s}$ are $\frac{3}{2}, \frac{4}{2}, \frac{5}{2}, \cdots, \frac{2 n-1}{2}$, which are only $(2 n-1)-2=2 n-3$ numbers. In this case, the set $A_{s}$ contains exactly $2 n-3$ distinct elements, which proves the conclusion.