Let $\sigma(n)$ denote the sum of positive divisors of a number $n$. A positive integer $N=2^r b$ is given, where $r$ and $b$ are positive integers and $b$ is odd. It is known that $\sigma(N)=2N-1$. Prove that $b$ and $\sigma(b)$ are coprime.
(J. Antalan, J. Dris)
Strange problem. If we denote $\sigma(2^r)=v$ and $2^{r+1}=u$ then one has $u\cdot b-v\cdot \sigma(b)=1$. This gives result. Even we do not need u and v are consequent numbers. Maybe something is wrong in the formulation.
this is copy paste from the official translation in English, of the 2017 contest in tex here (open with notepad)
In Russian, in tex the problems from 2017 are here