Without loss of generality assume that $a_1 < a_2 < \dots < a_n$. Consider these numbers on a circle, in the ascending order. All sums, other than $a_1 + a_n$ can be ordered as
$$a_1 + a_2 < a_2 + a_3 < \dots < a_{n-2} + a_{n-1} < a_{n-1} + a_n$$
We're done if there's no two sums that are equal. But if there is, the only sum that can be equal to another one is $a_1 + a_n$ so we would have $a_1 + a_n = a_{i-1} + a_i$ for some $1 < i < n$. Now let's replace $a_i$ and $a_{i+1}$.

$$a_n + a_1 = a_{i-1} + a_i < a_{i-1} + a_{i+1} < a_{i+1} + a_i < a_{i+2} + a_i.$$
So the new sums can be ordered as
$$a_1 + a_2 < a_2 + a_3 < \dots < a_{i-2} + a_{i-1} < a_n + a_1 < a_{i-1} + a_{i+1} < a_{i+1} + a_i < \dots < a_{n-1} + a_n.$$