If $a_1+b_1=m+1$ then $a_n+b_n=m+n$ $n^2+n=(a_1+a_2+...+a_n)+(b_1+b_2+...+b_n)=\frac{(m+n)(n+m+1)}{2}-\frac{m(m+1)}{2}=\frac{n(2m+1)+n^2}{2} \to n=2m-1$ Now take $(a_1,a_2,...,a_n)=(1,3,5,...,2m-1,2,4,...,2m-2)$ and $(b_1,b_2,...b_n)=(m,m-1,...,1,2m-1,2m-2,...,m+1)$ then $a_i+b_i=m+i$