Divide the positive integers into infinite sets

Question

I have a great problem from a past olympiad.

Divide the set of positive integers into two infinite sets, such that in each set the sum of any seven numbers is in the same set. Find all such partitions!

I need some help. Of course, I have found the trivial ones, such as even and odd numbers, but how to find them all? Thanks!

Answer 1

Let's assume we allow duplicates in the sum.

Let $\mathbb{N}=A\cup B$ with $1\in A$. Then we get $M_1:=\{1+6n|n\in \mathbb{N}\}\subset A$ and the sets $M_k:=\{k+6n|n\in \mathbb{N}\}; 2\leq k< 6$ are either subsets of $A$ or $B$ (Can easily be shown by contradiction using $x\in X\Rightarrow x+6nx\in X$ with $X\in\{A,B\}$, $n\in\mathbb{N}$ ).

Then it is only a short step to show, that $A$ is the set of all odd numbers: For $k,k+1\in A$ we get $A=\mathbb{N}$.

extra curl without duplicates

For $k\in A$ choose $a_1,\dots, a_6\in A$ such that $a_i>k+7$ and $a_i-1\in B$ (We can choose $a_i$ because $A,B$ are infinite). Then $A\ni k+a_1+\dots a_6 = (k+6)+(a_1-1)+\dots (a_6-1)$. Because of $a_i-1\in B$ we get $k+6\in A$, otherwise $k+a_1+\dots a_6\in B$.

Therefore $A,B$ are unions of the set $M_k$ and as before $A,B$ are the sets of even and odd numbers.