Slovenija 2008

Denote the elements of the set $K$ by $a_1,a_2,a_3,\dots$, so that $a_1<a_2<a_3<\dots$. We will prove the claim by induction. Obviously, the sum of all numbers less than $a_2$ is equal to $a_1$ and is therefore less than $a_2$.

Now, assume that for a positive integer $n$ we have $a_n>a_{n-1}+a_{n-2}+\cdots +a_2+a_1$. We show that a similar estimate holds for $n+1$.

We have $a_n<a_{n+1}$, so $a_{n+1}$ cannot be a divisor of $a_n$ and $a_n$ is a divisor of $a_{n+1}$. So, there exists a positive integer $k$ such that $a_{n+1}=k{a}_n$. Since $a_n$ and $a_{n+1}$ are not equal, we have $k\ge 2$ and $$a_{n+1}=k{a}_n\ge 2a_n=a_n+a_n>a_n+(a_{n-1}+a_{n-2}+\cdots +a_2+a_1)$$ by induction hypothesis. This inequality shows that the claim is true for $n+1$ and, by induction, for all positive integers.