Let $p_1,p_2,\cdots,p_k$ be any distinct prime numbers.
Then by the Chinese Remainder Theorem, the system of congruences
n + 1 & \equiv 0 \mod p_1^2 \\
n + 2 & \equiv 0 \mod p_2^2 \\
n + 3 & \equiv 0 \mod p_3^2 \\
n + k & \equiv 0 \mod p_k^2
has infinitely many solutions $n$. For any such solution $n$, we have that
$n + m$ is divisible by $p_m^2$ for $m=1,2,3,\cdots,k$, and so we have that $\mu(n+m)=0$ for $m=1,2,3\cdots,k$
We see that there are infinitely many natural numbers $n$ such that
In fact, if $k \geq 4$ and for some $n$ we have that
then it must in fact be the case that
This is because amongst the set of numbers
$$\lbrace n+1, n+2, n+3, n+4\rbrace$$
there will be a multiple of $4$. The Möbius function evaluated at this multiple of $4$ will be $0$.