Determine all natural numbers $n \geq 2$ for which there exist two arrangements $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ of the numbers $1,2, \ldots, n$ such that $\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots, a_{n}+b_{n}\right)$ are consecutive natural numbers.

(Walther Janous)

Answer. It works precisely when $n$ is odd.
We have

$$
\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\ldots+\left(a_{n}+b_{n}\right)=2(1+2+\ldots+n)=n(n+1) .
$$

On the other hand, there is a natural number $N$ such that

$$
a_{1}+b_{1}=N, a_{2}+b_{2}=N+1, \ldots, a_{n}+b_{n}=N+n-1
$$

and thus

$$
\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\ldots+\left(a_{n}+b_{n}\right)=n N+(1+\ldots+(n-1))=n N+n(n-1) / 2
$$

holds.

Consequently, $n(n+1)=n N+n(n-1) / 2$ and thus $N=n+1-\frac{n-1}{2}=\frac{n+3}{2}$. Therefore, $N$ is an integer if and only if $n$ is an odd natural number.

We now need to investigate whether for every odd number $n$ with $n \geq 3$ there are two suitable arrangements. Let $n=2 k+1$ with $k \geq 1$.

Experiments with $k=1$ and $k=2$ lead, for example, to the following pattern:

$$
\left(\begin{array}{cccccccc}
1 & k+2 & 2 & k+3 & 3 & \ldots & 2 k+1 & k+1 \\
k+1 & 1 & k+2 & 2 & k+3 & \ldots & k & 2 k+1
\end{array}\right)
$$

This results in the $2 k+1$ consecutive numbers $k+2, k+3, \ldots, 3 k+1,3 k+2$.

(Walther Janous)