66th Czech and Slovak Mathematical Olympiad

The considered fractions are symmetric in the sense that if we permute $(a,b,c)$ we get the same three fractions (possibly in different order). The same is true if we replace $(a,b,c)$ by $(-a,-b,-c)$. This simplifies the following casework.

Let us suppose that numbers $a,b,c$ are such that all three fractions are integers. If one of them equals zero, without loss of generality we assume $a=0$. Plugging this in we get that both $b/c$ and $c/b$ are integers. Hence $b$ and $c$ are nonzero and both $|b|\le |c|$ and $|c|\le |b|$ implying that $c=\pm b$. Moreover, $b=c$ is the denominator of the first fraction, hence $b+c\ne 0$ and $b=c$. Altogether, we obtain (clearly admissible) triplets $(0,c,c)$ and their permutations for any nonzero integer $c$.

It remains to solve the case $abc\ne 0$.

Due to the observations from the first paragraph we can assume that at least two of the numbers $a,b,c$ are positive. If all three were positive, the fraction with $\min \{a,b,c\}$ at its numerator lies between 0 and 1 and can't be integer.

Assume $a$ and $b$ are positive and $c=-d$ ($d>0$) negative. Plugging this in, the fractions rewrite as $$\frac{a}{d-b},\frac{b}{d-a},\frac{d}{a+b}$$ The last one implies that $d\ge a+b$. Hence the first one has positive denominator and since it is an integer, we have $a\ge d-b$, i.e. $d\le a+b$. Therefore $d=a+b$ ($c=-a-b$) and altogether we obtain triplets $(a,b,c)$ of nonzero numbers such that $a+b+c=0$. All such triplets are admissible as the value of each fraction then equals $-1$.