AUT_ABooklet_2023

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

We have $$(a_1+b_1)+(a_2+b_2)+\cdots +(a_n+b_n)=2(1+2+\cdots +n)=n(n+1).$$ 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 $$(a_1+b_1)+(a_2+b_2)+\cdots +(a_n+b_n)=nN+(1+\cdots +(n-1))=nN+\frac{n(n-1)}{2}.$$ 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: $$\left(\begin{array}{cccccccc}1& k+2& 2& k+3& 3& \dots & 2k+1& k+1\\ k+1& 1& k+2& 2& k+3& \dots & k& 2k+1\end{array}\right)$$ Summing the two rows gives the $2k+1$ consecutive numbers $k+2,k+3,\dots ,3k+1,3k+2$ as desired.

(Walther Janous) $\square$