we can clearly see that $n=2$ works with $\{a_1,a_2\}=\{1,3\}$ and we can check that $n\neq 3$ by checking the parity of $\{a_i +a_j | 1 \le i \le j \le 3 \}$ assume that there exist $n\in\mathbb{N}-\{1,2,3\}$ which there is $a_1,a_2,..,a_n\in\mathbb{N}$ such that $\{a_i +a_j | 1 \le i \le j \le n \}$ is CRS modulo $\frac{n(n+1)}{2}$ thus $a_i+a_j\not\equiv a_p+a_q(mod~\frac{n(n+1)}{2}),~~~ \forall i,j,p,q\in\{1,2,..,n\}$ which $\{i,j\}\neq \{p,q\}$ and $a_i\not\equiv a_j(mod~\frac{n(n+1)}{2}),~~~\forall i,j\in\{1,2,..,n\}$ which $i\neq j~~~~(*)$ let $a_i-a_j\equiv b_{(i,j)}(mod~\frac{n(n+1)}{2})~~~$ where $b_{(i,j)}\in\{1,2,..,\frac{n(n+1)}{2}\}~~~\forall i,j\in\{1,2,..,n\}$ which $i\neq j$ let $A=\{b_{(i,j)}|~i,j\in\{1,2,..,n\},~i\neq j\}\Rightarrow|A|=n(n-1) $ from $(*)$ we'll get that $b_{(i,j)}\neq b_{(p,q)},~~~\forall b_{(i,j)},b_{(i,q)}\in A,~(i,j)\neq(p,q)~~~(**)$ since $n> 3\Rightarrow 2n^2-2n>n^2+n\Rightarrow n^2-n>\frac{n(n+1)}{2}$ from pigeon hole principle there is at least $\lceil \frac{n(n-1)}{\frac{n(n+1)}{2}}\rceil \geq2$ members of $A$ with the same value which contradicts with $(**)$ $\therefore$ we can conclude that $\boxed{n=2}$ is the only solution $~~\square$ lmk if there is any mistakes