SOLVED
Let $\alpha\geq 1$. Is there a sequence of integers $n_k,m_k$ such that $n_k/m_k\to \alpha$ and $\sigma(n_k)=\sigma(m_k)$ for all $k\geq 1$, where $\sigma$ is the sum of divisors function?
Erdős
[Er74b] writes it is 'easy to prove the analogous result for $\phi(n)$'.
The answer is yes, proved by Pollack [Po15b].
