For $n = 2$, $n(n+1)/2 = 3$. Thus $a_1 = 1$, $a_2 = 2$ work.

Now suppose that $n \ge 3$ and that the set $\{a_1, \dots, a_n\}$ has the desired properties. We may assume without loss of generality, that $1 \le a_1 < a_2 < \dots < a_n \le \frac{n(n+1)}{2}$. Also let $a_{n+1} = a_1 + \frac{n(n+1)}{2}$.

First we note that for any two distinct pairs of integers $\{i, j\}$ and $\{k, \ell\}$, $i < j, k < \ell$, chosen from $\{1, 2, \dots, n+1\}$, $a_j - a_i \ne a_\ell - a_k$ for if they are equal, we would have $a_j + a_k = a_i + a_\ell$. In particular, the integers $a_{i+1} - a_i$, $i = 1, \dots, n$ are distinct. Since $a_{n+1} - a_1 = \frac{n(n+1)}{2}$, these differences are $1, 2, \dots, n$. This also means that $a_j - a_i \ge n+1$ if $j-i \ge 2$.

Let $t$ be the integer such that $a_{t+1} - a_t = 1$. Then since both $a_{t+2} - a_t$ and $a_{t+1} - a_{t-1}$ are $\ge n+1$, $a_{t+2} - a_{t+1} = a_t - a_{t-2} = n$ which is impossible unless $t=1$ or $t=n$. By symmetry, we can take $t=1$. Then, $a_2 - a_1 = 1$ and $a_3 - a_2 = n$. For $n=3$, this means $a_3 = a_2 + 3$. Also we have $n(n+1)/2 = 6$. Thus $a_2 + a_2 \equiv a_3 + a_3 \pmod 6$. So such a set does not exist for $n=3$.

Now let $n \ge 4$ and $s$ be such that $a_{s+1} - a_s = 2$. Then $s \ge 3$. By the same argument as before, we have $a_{s+2} - a_{s+1}$ and $a_s - a_{s-1}$ are both either $n$ or $n-1$. This is possible only when $s = 3$ or $n$. Thus either $a_5 - a_3 = n+1$ or $a_{n+1} - a_{n-1} = n+1$. Therefore one of these equals $a_3 - a_1 = n+1$ which is not possible. Thus the desired set of integers does not exist.