Junior Balkan Mathematical Olympiad

Suppose that $a+b+c+d=0$. Then $$abc+bcd+cda+dab=0(1)$$ If $abcd=0$, then one of the numbers, say $d$, must be $0$. In this case $abc=0$, and so at least one of the numbers $a,b,c$ will be equal to $0$, making one of the given equations impossible. Hence $abcd\ne 0$ and, from (1), $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0.$$ implying $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}.$$ It follows that $(a+b)(b+c)(c+a)=0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $2+3=0$, a contradiction). Thus $a+b+c+d\ne 0$.