Let $\sigma_k(n)=\sum_{d|n} d^k,$ for a positive integer $n$ and $k\geq 0$. A lot is known about the averages for the functions $\sigma_k(n),$ such as the estimates $$ \sum_{n\leq x} \sigma_0(n)=x \log x + (2 \gamma -1) x +{\cal O}(\sqrt{x}), $$ and $$ \sum_{n\leq x} \sigma_1(n)=\frac{\pi^2}{12}x^2+E_1(x), $$ where to the best of my knowledge, the error term is $E_1(x)=\Omega(x \log\log x).$
Now let $N$ be large. Is it possible to bound from below the signed sum $$ S_N(\varepsilon_1,\cdots,\varepsilon_N):=\left| \sum_{n=1}^{N} \varepsilon_n \left\{N \sigma_0(n)-\sigma_1(n) \right)\}\right| $$ for all or almost all $(\varepsilon_1,\cdots,\varepsilon_N)\in \{\pm 1\}^N$ by an increasing function of $N$?
Alternatively, perhaps it is possible to bound from below the following quantity $$ \max_{N'\leq N} S_{N'}(\epsilon_1,\ldots,\epsilon_{N'})? $$ and see how it varies with increasing $N.$
Related Question: In a related question I had asked here about $X_0(\varepsilon)$, Greg Martin used a nice argument to show that the sum can actually be made constant in absolute value for some sign pattern $\varepsilon.$ Can that argument work here?
Essentially for almost all sign patterns $\varepsilon$, $X_0(\varepsilon)$ will have order of magnitude $\sqrt{N} (\log N)^{3/2}$ thus $NX_0(\varepsilon)$ will have order of magnitude $N^{3/2}N (\log N)^{3/2}$ but will have support on $N \mathbb{Z}$.
So manipulating this sum by changing the signs of the pattern $\epsilon$ on primes $p\leq N,$ can only change the sum by a multiple of $N$. Thus it seems like this approach cannot in general give a lower bound better than $\Omega(N).$ Even then, the dependence between $\sigma_0(n)$ and $\sigma_1(n)$ complicates things.
I am hoping experts in number theory may see a more sophisticated approach to a lower bound.
Discussion: Define the uniformly distributed random variables $\varepsilon = (\varepsilon_1,\dots,\varepsilon_N) \in \{\pm1\}^N$. Then, the random variables $$ X_0(\varepsilon) = \sum_{n=1}^N \varepsilon_n \sigma_0(n) $$ and $$ X_1(\varepsilon) = \sum_{n=1}^N \varepsilon_n \sigma_1(n) $$ are both zero mean with variances $$ \textrm{Var}(X_0(\varepsilon)) = \sum_{n=1}^N \sigma_0(n)^2 \sim \frac{N(\log N)^3}{\pi^2}, $$ and $$ \textrm{Var}(X_1(\varepsilon)) = \sum_{n=1}^N \sigma_1(n)^2 \sim \frac{5}{6}\zeta(3)N^3, $$ respectively. The quantity I am interested in can be represented by the random variable $$ Z(\varepsilon):=N X_0(\varepsilon)-X_1(\varepsilon) $$ which is zero mean with variance roughly $O(N^3)$ (the two random variables $X_i$ are not independent, however). This means the typical value of the quantity is $O(N^{3/2}).$
Can the approach in the answer the linked question be modified to work here?
