Let $\mu(n)$ be the Mobius function. Let us define $\mu^+(n)$ to be $\mu(n)$ if $\mu(n)>0$ and $0$ otherwise. Is there a known asymptotic formula for $$ \sum_{n \leq N} \mu^+(n), $$ and similarly for $$ \sum_{ \substack{n \leq N \\ n \equiv a (\mod q)}} \mu^+(n). $$ I would greatly any appreciate references or comments. Thank you very much.

7$\begingroup$ I think the results you seek probably follow easily from the identity $\mu^{+}(n) = \frac{\mu(n)^{2} + \mu(n)}{2}$ combined with known results about $\sum_{n \equiv a \pmod{q}, n \leq x} \mu(n)$, and the count of the number of squarefree integers in arithmetic progressions. $\endgroup$ – Jeremy Rouse Jun 1 '17 at 14:24
I'll have a stab at this. The relation $$ \sum_{n\leq X:n\equiv a~(mod~q)} \mu^2(n)=\frac{6}{\pi^2} \prod_{pq} \left(1\frac{1}{p^2} \right)\frac{X}{q}+E(X,q,a) $$ where the error term $E$ is $O_{\varepsilon}\left(\sqrt{X/q} +q^{\frac{1}{2}+\varepsilon}\right) $ provided $q\leq X^{\frac{2}{3}\varepsilon},$ together with some recent results (assuming GRH) of the form
$$ \sum_{n\leq X:n\equiv a~(mod~q)} \mu(n) \ll_{\varepsilon} \sqrt{X} \exp\left( (\log X)^{3/5} (\log \log X)^{(16/5)+\varepsilon} \right) $$
(see the paper on arxiv here) and the observation $\mu^{+} (n)=\frac{1}{2}(\mu(n)+\mu^2(n))$ suffices to yield an answer.
Edit The paper here gives the bound $$ O\left( \frac{X}{q\log^C X}\right) $$
for any $C>0$, for $X$ large enough.