Bulgaria MO 2005 P3

Conclusion: It does not exist such set $M$. Suppose that there exist such $M$. It's easy to see that $M$ is infinite and we can assume that $1\not \in M$, lets say $\min_{m\in M}m\geq 2$. If there exists $x,y\in M$ such that $x<y<2x$ then the rational number $r=\frac{1}{x}-\frac{1}{y}$ is greater than $0$ and less than $1$. Therefore $r$ must be written as sum of reciprocals of finite numbers in $M$ (say $\frac{1}{y_1}+\dots+\frac{1}{y_n}$). Notice that none of the numbers $y_i$ is $x$ nor $y$ so $\frac{1}{x}$ can be written as $\frac{1}{x}$ and $\frac{1}{y}+\frac{1}{y_1}+\dots+\frac{1}{y_n}$ at the same time, a contradiction! Thus the elements of $M$ must form a sequence $\{u_n\}: u_1\geq 2$ and $u_n\geq 2u_{n+1}$. We have: $$\sum_{i=1}^{\infty}\frac{1}{u_i}\leq \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots=1$$So we must have $u_n=2^n$ for $n=1, 2, \dots$ otherwise there exists a rational number $\epsilon>0$ such that $1-\epsilon>$ sum of reciprocals of all elements in $M$. However we have known that $\frac{1}{3}=\overline{0.010101\dots}_2$, that leads to a contradiction!