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.