I claim the answer are all odd $n$. First, let $n$ be such that such two permutations exist. Let $a_i+b_i=k+i$ for $1\le i\le n$. We have that $\textstyle \sum_{1\le i\le n}(a_i+b_i) = kn + n(n+1)/2$. On the other hand, $\textstyle \sum_{1\le i\le n}(a_i+b_i) = 2\cdot (1+\cdots +n)$, yielding $k=(n+1)/2$. Thus, such an $n$ exists only if $2\mid n+1$, i.e., $n$ is odd. As for the example, set $n=2t+1$. Observe that $(a_1,\dots,a_n)=(1,\dots,n)$, $(b_1,\dots,b_n)=(k+1,k+2,\dots,2k+1,1,2,3,\dots,k)$ satisfy the condition.

Suppose that $a_1+b_1 < a_2+b_2 <\ldots < a_n+b_n$ (otherwise just relabel) and let $d+1=a_1+b_1$, so $a_i+b_i=d+i$. So we get $$\sum_{i=1}^n(a_i+b_i)=nd+\frac{n(n+1)}{2}=\sum_{i=1}^n a_i+\sum_{i=1}^n b_i=n(n+1)\Rightarrow d=\frac{n+1}{2}$$ So $n=2k-1$ must be odd. And here is the construction: $$(a_1,a_2,\ldots,a_n)=(1,k+1,2,k+2,\ldots,2k-1,k)$$$$(b_1,b_2,\ldots,b_n)=(k,1,k+1,2,\ldots,k-1,2k-1)$$where the even entries are consecutive numbers and so are the odd ones. $\square$

$$\sum_{i=1}^n(a_i+b_i)=n*(a_1+b_1)+n(n-1)/2=n(n-1)$$So $a_1+b_1=(n-1)/2$ , the construction is already given by the other people so I'm not gonna bother.