Bariz - Problem

I'll show it is odd. Group equal $ a_i$ to obtain a sum of the form $ \sum \frac {n_i}{b_i}$ with $ \sum n_i = 10$ and $ \{a_1,...,a_{10}\} = \{b_1,...,b_r\}$. Then the number of such solutions is $ \frac {10!}{n_1!\cdots n_r!}$. To check parity, we only need to consider arrangements where this number is odd. But $ \frac {10!}{n_1!\cdots n_r!}$ is odd only if $ n_1 = 2,n_2 = 8$, or $ n_1 = 10$. In the first we have $ 1 = \frac {2}{a} + \frac {8}{b} \Leftrightarrow (a - 2)(b - 8) = 16$, which has solutions $ (a,b)\in\{(3,24),(4,16),(6,12),(18,9)\}$, each appearing $ 45$ times, $ 4\cdot 45$ is even. In the latter case we have one solution. So the total is is odd.