, 2019

Answer: 2.
It is easy to verify that for any odd prime number $p$ we have $s(p) = \sigma(p) = 2$ and hence $2s(n) - \sigma(n) = 2$. We will show that $2s(n) - \sigma(n) \le 2$ for every positive integer $n$. Let $1 = d_1 < d_2 < \dots < d_k = n$ be positive divisors of $n$. It is well known that $d_i d_{k+1-i} = n$ for $1 \le i \le k$. If $d_i + 1|n+1$, then $d_i + 1|d_i d_{k+1-i} + 1 - (d_i + 1) = d_i(d_{k+1-i} - 1)$. $\Rightarrow d_i|d_{k+1-i} - 1$ since $(d_i + 1, d_i) = 1$. $\Rightarrow i = k$ or $i < k+1-i$. $\Rightarrow i = k$ or $i \le k-i$. $\Rightarrow s(n) \le 1 + \frac{k}{2} \Rightarrow 2s(n) \le k+2$ and the result follows.