only 3 and 4, im to lazy to post my prove.

Why not just say, for what values of n can we find n numbers so that their pairwise sums are n(n-1)/2 consecutive integers?

For $n=3$ we have $1,2,3$ and for $n=4$ we have $1,2,3,5$ For $n>5$, let $n$ distinct integers are $a_1,a_2,...,a_n$ Suppose that $\binom{n}{2}$ consecutive sums are $t+1,t+2,...,t+\binom{n}{2}$ Easy to see that $a_1+a_2=t+1,a_1+a_3=t+2,a_{n-2}+a_n=t+\binom{n}{2}-1,a_{n-1}+a_n=t+\binom{n}{2}$ So $a_3=a_2+1,a_{n-1}=a_{n-2}+1$, we get $a_3+a_{n-2}=a_2+a_{n-1}$, Since there are $\binom{n}{2}$ ways to choose $2$ distinct numbers from $a_1,a_2,...,a_n$ and there are $\binom{n}{2}$ distinct sums When $n>5$ we get that all of $a_3,a_{n-2},a_2,a_{n-1}$ are all distinct, contradiction. We left the case where $n=5$, we have $a_3=a_2+1,a_4=a_3+1$, let $a_3=x$ So $2x-1,2x,2x+1 \in \{ t+1,t+2,...,t+10\}$ and the rest is just case bash.

There is a long solution by considering cases on $a_2-a_1$ and achieving contradictions for each; I'll leave that solution for the viewer to find. Instead, here's the slick solution: The answer is $n = 3, 4$ only. For $n \geq 6$, consider the chain of inequalities $$a_1+a_2<a_1+a_3<\cdots<a_{n-2}+a_{n-1}<a_{n-1}+a_n.$$Notice that the condition implies $a_3-a_2=a_{n-1}-a_{n-2}=1$, or $a_3+a_{n-2}=a_2+a_{n-1}$. Thus two sums are equal, which is a contradiction. It suffices to consider $n=3, 4, 5$. 3 works; choose 1, 2, 3, and 4 works by choosing 1, 2, 3, 5. When $n=5$, note that $a_2+2=a_3+1=a_4$, so as a result, $$a_1+a_2<a_1+a_3<a_1+a_4.$$The next term in the chain of consecutive sums must be $a_2+a_3$, so $a_2 = 2+a_1$. Similarly, $a_4+2=a_5$, so $a_1+a_5=a_2+a_4$, which is also a contradiction. We are done.