= Solution
The <Möbius function> is defined by $\mu(1)=1$ and, for $n>1$,
$$
\mu(n)=
\begin{cases}
0,&p^2\mid n\text{ for some prime }p,\\
(-1)^r,&n\text{ is a product of }r\text{ distinct primes}.
\end{cases}
$$
It is a <multiplicative arithmetic function>: $\mu(mn)=\mu(m)\mu(n)$ whenever $m$ and $n$ are <coprime integers>[coprime].
The summand $\mu(d)^2/\phi(d)$ is multiplicative, so its <divisor sum> is multiplicative. For a <prime power> $p^a$ with $a\geq1$, only $d=1,p$ contribute and
$$
\sum_{d\mid p^a}\frac{\mu(d)^2}{\phi(d)}
=1+\frac1{p-1}=\frac p{p-1}
=\frac{p^a}{\phi(p^a)}.
$$
Multiplying these local identities over the prime divisors of $n$ gives
$$
\boxed{\sum_{d\mid n}\frac{\mu(d)^2}{\phi(d)}=\frac n{\phi(n)}}.
$$
For the final claim, choose distinct primes $p_0,\ldots,p_k$. The moduli $p_j^2$ are pairwise coprime, so the <Chinese remainder theorem> gives an integer $a$ satisfying
$$
a\equiv-j\pmod{p_j^2},\qquad 0\leq j\leq k.
$$
Every positive integer $n\equiv a\pmod{\prod_jp_j^2}$ then has $p_j^2\mid n+j$. Hence
$$
\boxed{\mu(n)=\mu(n+1)=\cdots=\mu(n+k)=0}.
$$
There are infinitely many positive representatives of this congruence class.
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