62nd Czech and Slovak Mathematical Olympiad

Initially we will prove that the set $A$ consists of at most two numbers. Suppose that three numbers $a<b<c$ belong to the set $A$. Then the numbers $a+b<a+c<b+c$ are in $B$ and therefore the number $$\frac{b+c}{a+c}=1+\frac{b-a}{a+c}$$ has to be in $A$. This is a contradiction because $0<b-a<a+c$ and the number is not integer.

If the set $B$ contains four numbers $k<l<m<n$, then the set $A$ will contain three different numbers $n/k$, $n/l$, $n/m$. So the set $B$ has at most three elements and $A\cup B$ has at most five elements.

We achieve the number $5$ of elements if $A=\{a,b\}$, $B=\{k,l,m\}$, where $a<b$ and $l/k=m/l=a$, $m/k=b$. Then $b=a^2$ ($a\ge 2$) and $a+a^2$ is one of the elements of $B$; the next two elements are either $a^2+a^3$ and $a^3+a^4$ or $1+a$ and $a^2+a^3$. E.g. sets $A=\{2,4\}$, $B=\{3,6,12\}$ have five elements together.