Natural numbers $n$ satisfying $\mu(n+1)+\mu(n+2)+\cdots+\mu(n+2019)=2019$

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Determine all nautral numbers $n$ satisfying $$\mu(n+1)+\mu(n+2)+\cdots+\mu(n+2019)=2019$$

Where $\mu (n) = (-1)^{\omega(n)}$ if $n$ is square free, and $\mu(n)=0$ otherwise. $w(n)$ denotes distinct prime divisors of $n$. For example. $w(6) = 2, \mu(6)=1$

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Four consecutive integers include one divisible by $4$, therefore no such $n$ exists.