59th Ukrainian National Mathematical Olympiad

Let us first find rational numbers satisfying the equality, and then multiply them by the least common multiplier of the denominators in order to obtain integers. One can fix $b_1=-2^{n-1}$, $b_2=-2^{n-2}$, ..., $b_{n-1}=-2^1$. Let us rewrite the equation as follows: $$\frac{b_1}{b_1}+\frac{b_1}{b_2}+\frac{b_1}{b_3}+\cdots +\frac{b_1}{b_{n-1}}-\frac{b_2}{b_1}-\frac{b_2}{b_2}-\frac{b_2}{b_3}-\cdots -\frac{b_2}{b_{n-1}}=\frac{b_2}{b_n}-\frac{b_1}{b_n}\text{or}$$ $$X=\frac{2^{n-1}}{2^{n-2}}-\frac{2^{n-2}}{2^{n-1}}+\frac{2^{n-1}}{2^{n-3}}-\frac{2^{n-2}}{2^{n-3}}+\frac{2^{n-1}}{2^{n-4}}-\frac{2^{n-2}}{2^{n-4}}+\cdots +\frac{2^{n-1}}{2^1}-\frac{2^{n-2}}{2^1}=\frac{2^{n-1}}{b_n}-\frac{2^{n-2}}{b_n}\Rightarrow$$ $$X=2-\frac{1}{2}+2^2-2+2^3-2^2+\cdots +2^{n-2}-2^{n-3}=2^{n-2}-\frac{1}{2}=\frac{2^{n-1}-1}{2}=\frac{2^{n-2}}{b_n}\Rightarrow$$ $$b_n=\frac{2^{n-1}}{2^{n-1}-1}>0.$$

All $b_k$, $k=\overline{1,n-1}$ are negative, and from the last inequality it follows that $b_1<b_2<\cdots <b_{n-1}<b_n$. Finally, we can get aforementioned integers as follows: $$a_k=b_k\cdot (2^{n-1}-1),k=\overline{1,n}.$$