Let $\mu$ be the Möbius function. Let $$\lambda_d = \begin{cases} \frac{\log D/d}{\log D} \mu(d)&\text{for $d\leq D$,}\\ 0 &\text{otherwise.}\end{cases}$$ (Selberg's weights also work.) Then it is well-known that, for $N\ggg D^2$, $$\sum_{n\leq N} \left(\sum_{d|n} \lambda_d\right)^2 = \frac{N}{\log D} + \text{lower-order terms}.$$
What is the order of $$S_1 = \sum_{n\leq N} \left|\sum_{d|n} \lambda_d\right|,$$ under the same assumptions? Of course we can give an upper bound $S_1\leq N/\sqrt{\log D} + \dots$ from the above by Cauchy-Schwarz, but can one do better? Is $S_1$ perhaps $O(N/\log D)$? (It certainly is if we take away the absolute value.)
