Solution. We will use method from contradiction. Suppose $a + b + c + d = 0$. Thus $a + b = -(c + d)$. Therefore, we can obtain such equalities:
$$(a + b)^3 = -(c + d)^3 \Leftrightarrow a^3 + b^3 + 3ab(a + b) = -c^3 - d^3 - 3cd(c + d) \Leftrightarrow \\ 3ab(a + b) + 3cd(c + d) = 0 \Leftrightarrow -3ab(c + d) - 3cd(a + b) = 0 \Leftrightarrow \\ ab(c + d) + cd(a + b) = 0 \Leftrightarrow abc + abd + acd + bcd = 0.$$
That contradicts the conditions since:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \neq 0 \Leftrightarrow \frac{abc + abd + acd + bcd}{abcd} \neq 0 \Leftrightarrow abc + abd + acd + bcd \neq 0.$$