Cute Yes we claim that such fractions do exist . Consider the following 2018 fractions. $\frac {1+x}{x}, \frac {2+x}{2x}, \frac {3+x}{3x}, \dots \frac {2018+x}{2018x}$ . Choose $x$ in such a way such that all the fractions are irreducible . $x=(2018)!+1$ works ,for instance. As $(x,i)=1 $ $\forall 1 \leq i \leq 2018$ Now the difference of any 2 fractions is $$\frac {i+x}{ix}-\frac {j+x}{jx}=\frac {j-i}{ij}.$$ Since $ij < x $ $\forall 1 \leq x,y \leq 2018$ ,we are done.