, 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 $a_n > a_{n-1} + a_{n-2} + \dots + a_2 + a_1$. 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
$$a_{n+1} = k a_n \ge 2 a_n = a_n + a_n > a_n + (a_{n-1} + a_{n-2} + \dots + a_2 + a_1)$$
by induction hypothesis. This inequality shows that the claim is true for $n+1$ and, by induction, for all positive integers.