Seventeenth Stars of Mathematics Competition

The required integers are $n=3$ and $n=4$. In the former case, let $(a_1, a_2, a_3) = (1, 2, 3)$; the sums of pairs form the arithmetic sequence $3, 4, 5$. In the other case, let $(a_1, a_2, a_3, a_4) = (1, 3, 4, 5)$; the sums of pairs form the arithmetic sequence $4, 5, 6, 7, 8, 9$.
Since the $n$ numbers are pairwise distinct, so are the $\frac{1}{2}n(n-1)$ sums of pairs — otherwise, the difference of the arithmetic sequence would be zero, so the first and the second smallest sums would be equal, hence the second and the third smallest numbers would be equal, and we would reach a contradiction.

Let now $n \ge 5$, and suppose, if possible, that $a_1, a_2, \dots, a_n$ satisfy the condition in the statement. Without loss of generality, we may and will assume that $a_1 < a_2 < \dots < a_n$. Let $d$ be the difference of the corresponding arithmetic sequence of sums.
The smallest sum is $a_1 + a_2$, and the second smallest sum is $a_1 + a_3$, so $a_3 - a_2 = d$. The largest sum is $a_{n-1} + a_n$, and the second largest sum is $a_{n-2} + a_n$, so $a_{n-1} - a_{n-2} = d$. Consequently, $a_2 + a_{n-1} = (a_3 - d) + (a_{n-2} + d) = a_3 + a_{n-2}$. If $n \ge 6$, the leftmost sum and the rightmost sum correspond to distinct pairs, so they are at least $d$ distance apart. This contradiction forces $n=5$.
Let $n=5$. By the preceding, $a_3 - a_2 = d$, and $a_4 - a_3 = d$, so $2a_3 = a_2 + a_4$. To reach a contradiction, we present two approaches.

1st Approach. Evaluate the sum $s = \sum_{1 \le i < j \le 5} (a_i + a_j)$ in two different ways. On the one hand, $s = 5(a_1 + a_2 + a_4 + a_5)$. On the other hand, $s = 4(a_1 + a_2 + a_3 + a_4 + a_5)$, since each $a_k$ occurs in exactly four sums $a_i + a_j$, $i < j$. Equate the two and clear out like terms, to get $a_1 + a_2 + a_4 + a_5 = 4a_3 = 2(a_2 + a_4)$. Consequently, $a_1 + a_5 = a_2 + a_4$ which is the desired contradiction.

2nd Approach. Notice that the third smallest sum is $a_1 + a_4$ which is $d$ larger than $a_1 + a_3$, and the third largest sum is $a_2 + a_5$. Between these lie $a_1 + a_5$ and the consecutive sums $a_2 + a_3 < a_2 + a_4 < a_3 + a_4$. Then $a_1 + a_5$ is either the fourth smallest sum or the fourth largest.
Without loss of generality, we may and will assume that $a_1 + a_5$ is the fourth smallest sum. The string of ten sums is then $a_1 + a_2 < a_1 + a_3 < a_1 + a_4 < a_1 + a_5 < a_2 + a_3 < a_2 + a_4 < a_3 + a_4 < a_2 + a_5 < a_3 + a_5 < a_4 + a_5$.
Evaluate $a_5 - a_4$ in two different ways. On the one hand, $a_5 - a_4 = (a_1 + a_5) - (a_1 + a_4) = d$, by the third inequality from the left. On the other hand,
$$a_5 - a_4 = ((a_2 + a_5) - (a_3 + a_4)) + (a_3 - a_2) = d + d = 2d,$$
by the third inequality from the right and $a_3 - a_2 = d$. This is the desired contradiction.