From the first equality we can get that $a-b = \frac{1}{c} - \frac{1}{b}$ or that $(a-b) = \frac{b-c}{bc}$.
Analogously $b-c = \frac{c-a}{ac}$ and $c-a = \frac{a-b}{ab}$. Thus we obtain: $a-b = \frac{a-b}{(abc)^2}$. Since $a$, $b$ are distinct, $(abc)^2 = 1$, which means that $abc = \pm 1$. Let us show that there exist numbers $a$, $b$, $c$ for which both of the values can be achieved.

Take $a = 1$. Then we have two equations on $b$, $c$: $1 + \frac{1}{b} = c + 1$ and $1 + \frac{1}{b} = b + \frac{1}{c}$.
From the first one we get that $c = \frac{1}{b} \Rightarrow 2b = 1 + \frac{1}{b}$ or $2b^2 - b - 1 = 0$. This equation has two roots $-b_1 = -\frac{1}{2}$ and $b_2 = 1$. The second root coincides with $a$, thus we take $b = -\frac{1}{2}$ and $c = -2$. For these $a$, $b$, $c$ we have $abc = 1$ and it is easy to check that they satisfy the problem conditions.

Now take $a = -1 \Rightarrow -1 + \frac{1}{b} = c - 1$ and $-1 + \frac{1}{b} = b + \frac{1}{c} \Rightarrow c = \frac{1}{b}$ and $2b = \frac{1}{b} - 1$. So $2b^2 + b - 1 = 0$. The roots of this equation are $b_1 = \frac{1}{2}$ and $b_2 = -1$. Again we choose only one value $b = \frac{1}{2} \Rightarrow c = 2$ and $abc = -1$. A direct check shows that these $a$, $b$, $c$ fulfill the conditions.