Mathematica competitions in Croatia

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!