a) We will show that for four given numbers one can always find two numbers on the board such that one is not more than twice the other one. Suppose by contradiction, it is not true. Let the given numbers be $a < b < c < d$. Then $d + \frac{b}{a} > c + \frac{b}{a} \Rightarrow d + \frac{b}{a} > 2(c + \frac{b}{a}) \Rightarrow d + \frac{b}{a} > 1$.

On the other hand, $d + \frac{b}{c} > d + \frac{a}{c} \Rightarrow d + \frac{b}{c} > 2(d + \frac{a}{c}) \Rightarrow d < \frac{b}{c} < 1$. Thus, we get a contradiction.

b) We will give an example of three numbers for which the condition doesn't hold: $a = 10^{-1}$, $b = 10^{-3}$ and $c = 10^{-9}$. Then, the following numbers are written on the board:

$10^{-1} + 10^{6}$, $10^{-1} + 10^{-6}$, $10^{-3} + 10^{8}$, $10^{-3} + 10^{-8}$, $10^{-9} + 10^{2}$, $10^{-9} + 10^{-2}$.