Croatian Mathematical Competitions

On the contrary, let's assume that all sides are of different lengths; i.e. $a < b < c < d$ ($a$, $b$, $c$, $d$ are lengths of the sides, ordered by their length). Each of the lengths is a divisor of $S = a + b + c + d$, by assumption.
Also, it must be $a + b + c > d$ or equivalently $S > 2d$ and finally $\frac{S}{d} > 2$. Since $d$ is a divisor of $S$ it follows that $\frac{S}{d} \ge 3$. Now we have $c < d \le \frac{S}{3}$ and since $c$ is a divisor of $S$, it follows that $c \le \frac{S}{4}$. Analogously we obtain $b \le \frac{S}{5}$ and $a \le \frac{S}{6}$.
Then
$$S = a + b + c + d \le \frac{S}{6} + \frac{S}{5} + \frac{S}{4} + \frac{S}{3} = S \left( \frac{1}{6} + \frac{1}{5} + \frac{1}{4} + \frac{1}{3} \right) = \frac{57}{60} S < S.$$
Contradiction!