Let $S$ be a set of $n$ distinct real numbers. Let $A_{S}$ be the set of numbers that occur as averages of two distinct
elements of $S$. For a given $n \geq 2$, what is the smallest possible number of elements in $A_{S}$?
1. Let \( S = \{a_1, a_2, \ldots, a_n\} \) be a set of \( n \) distinct real numbers such that \( a_1 < a_2 < \cdots < a_n \).
2. We need to find the smallest possible number of elements in \( A_S \), where \( A_S \) is the set of numbers that occur as averages of two distinct elements of \( S \).

3. Consider the averages of all pairs of distinct elements in \( S \). The average of \( a_i \) and \( a_j \) (where \( i < j \)) is given by:
   \[
   \frac{a_i + a_j}{2}
   \]

4. We need to determine the number of distinct values that these averages can take. To do this, we first consider the sums \( a_i + a_j \) for \( 1 \leq i < j \leq n \).

5. The smallest sum is \( a_1 + a_2 \) and the largest sum is \( a_{n-1} + a_n \). The sums \( a_i + a_j \) for \( 1 \leq i < j \leq n \) form a sequence:
   \[
   a_1 + a_2, a_1 + a_3, \ldots, a_1 + a_n, a_2 + a_3, \ldots, a_{n-1} + a_n
   \]

6. To find the number of distinct averages, we need to count the number of distinct sums \( a_i + a_j \). If \( S \) is an arithmetic sequence, the sums \( a_i + a_j \) will also form an arithmetic sequence.

7. Suppose \( S \) is an arithmetic sequence with common difference \( d \). Then:
   \[
   a_i = a_1 + (i-1)d
   \]
   The sum of two elements \( a_i \) and \( a_j \) (where \( i < j \)) is:
   \[
   a_i + a_j = (a_1 + (i-1)d) + (a_1 + (j-1)d) = 2a_1 + (i+j-2)d
   \]

8. The distinct sums \( a_i + a_j \) will be of the form \( 2a_1 + kd \) where \( k \) ranges from \( 1 \) to \( 2n-3 \). Therefore, there are \( 2n-3 \) distinct sums.

9. Since each distinct sum corresponds to a distinct average, the number of distinct averages is also \( 2n-3 \).

10. Therefore, the smallest possible number of elements in \( A_S \) is \( 2n-3 \).

The final answer is \( \boxed{ 2n-3 } \).