The permutations exist if and only if $n$ is odd.

We have
$$(a_1 + b_1) + (a_2 + b_2) + \dots + (a_n + b_n) = 2(1 + 2 + \dots + n) = n(n+1).$$
On the other hand, there is a natural number $N$ such that
$$a_1 + b_1 = N,\ a_2 + b_2 = N + 1,\ \dots,\ a_n + b_n = N + n - 1$$
and therefore
$$(a_1 + b_1) + (a_2 + b_2) + \dots + (a_n + b_n) = nN + (1 + \dots + (n-1)) = nN + \frac{n(n-1)}{2}.$$
We obtain the equation $n(n+1) = nN + n(n-1)/2$ which becomes $N = n+1 - \frac{n-1}{2} = \frac{n+3}{2}$.
Therefore, the number $N$ is an integer if and only if $n$ is odd.

It remains to investigate if two permutations with the desired property exist for every odd number $n$ with $n \ge 3$. Let $n = 2k + 1$ with $k \ge 1$.
Experimenting with $k=1$ and $k=2$ can lead to the following pattern:
$$\begin{pmatrix} 1 & k+2 & 2 & k+3 & 3 & \dots & 2k+1 & k+1 \\ k+1 & 1 & k+2 & 2 & k+3 & \dots & k & 2k+1 \end{pmatrix}$$
Summing the two rows gives the $2k+1$ consecutive numbers $k+2, k+3, \dots, 3k+1, 3k+2$ as desired.

(Walther Janous) $\square$