Junior Balkan Mathematical Olympiad

Suppose that $a + b + c + d = 0$. Then
$$abc + bcd + cda + dab = 0 \quad (1)$$
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 $abcd \neq 0$ 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 $a + b + c + d \neq 0$.