In Apostol's Introduction to Analytic Number Theory he uses $\sum_{q to prove this bound as Theorem 3.4. We may be able to do better by using the exact expression $ \begin{align} \sum_{q where $\{x\}$ is the fractional part of $x$.
Then proceeding in the same way $ \begin{align} S(n) = \sum_{k\le n}\sigma(k) & = \sum_{d\le n}\sum_{q\le n/d}q \\ & = \sum_{d\le n}\left(\frac{n^2}{2d^2}+\frac{n}{2d}-\frac{n\{n/d\}}{d}+O(1)\right) \\ & = \frac{1}{2}n^2\zeta(2)+\frac{1}{2}n\log(n)-n\sum_{q\le n}\frac{\{n/d\}}{d}+O(n) \end{align} $
Write $t(n)=n\sum\{n/d\}/d$ for the third term. Since $|\{n/d\}|<1$, $t(n)=O(n\log(n))$.
Let $S(n)=n^2\pi^2/12+F(n)$. Gronwall's Theorem gives us that there are arbitrarily large $N$ with $\sigma(N)>N\log\log(N)$. For such a $N$ we have $ \begin{align} S(N)-S(N-1) & =\sigma(N)>N\log\log(N) \\ F(N)-F(N-1) & > N \log\log(N)-(2N-1)\pi^2/12 \\ \max(|F(N)|,|F(N-1)|) & > N\log\log(N)/2 - N \end{align} $ So $F(n)=\Omega(n\log\log(n))$. But $F(n) = n\log(n)/2-t(n)+O(n)$.
Write $t(n)=An\log(n)-f(n)$ with $f(n)=o(n\log(n))$. Then either:
$A\neq \frac{1}{2}$ and $F(n) = \left(\frac{1}{2}-A\right)n\log(n)+o(n\log(n))$
or
$A = \frac{1}{2}$ and $F(n) = f(n) + O(n)$, with $f(n)=o(n\log(n))$ and $f(n)=\Omega(n\log\log(n))$
Unfortunately I can't resolve which case applies, and in fact, if it's the second then I haven't really answered the question after all.