74th Romanian Mathematical Olympiad

Let $a=\frac{p}{q}$, where $p$ and $q$ are non-zero natural numbers, relatively prime, with $p<q$. Then $0=\{qa\}\ge \{qb\}\ge 0$, thus $qb$ is a natural number. It follows that $b=\frac{s}{q}$, where $s$ is a nonzero natural number.

Considering $n=q-1$ in the relation from the hypothesis, we get that $\{-\frac{p}{q}\}\ge \{-\frac{s}{q}\}$, that is $1-\frac{p}{q}\ge 1-\frac{s}{q}⟺p\le s$. So $p=s$, hence the requirement of the problem.

Alternative solution. Analogously to solution 1, we have $a=\frac{p}{q}$, $b=\frac{s}{q}$, $0<s\le p<q$, $(p,q)=1$. Then there is a natural number $k$ such that $kp\equiv 1(\mathrm{mod}q)$. For this $k$ we have $\{ka\}=\{k\cdot \frac{p}{q}\}=\frac{1}{q}$. Then $\frac{1}{q}=\{ka\}\ge \{kb\}=\{k\cdot \frac{s}{q}\}$, from which it follows that the number $\{k\cdot \frac{s}{q}\}$ is equal to either $0$ or $\frac{1}{q}$. If $\{k\cdot \frac{s}{q}\}=0$, then $q\mid ks$; since $(q,k)=1$, it means that $q\mid s$, false. If $\{k\cdot \frac{s}{q}\}=\frac{1}{q}$, then $ks\equiv 1(\mathrm{mod}q)$, therefore $q\mid k(p-s)$; since $(q,k)=1$, it means that $q\mid p-s$. But $0\le p-s<q$, therefore $p-s=0$. It turns out that $p=s$, therefore $a=b$.