One more of those

A compressed solution for those who are in a hurry:

We note that $f(n)=\sum_{d|n}\mu(d)\phi(d)$ is a multiplicative function. For a prime power $p^m$ we have $f(p^m)=\mu(1)\phi(1)+\mu(p)\phi(p)+0=2-p$ since $\mu(p^k)=0$ for $k>1$ . Thus $f(n)=\prod_{p|n}(2-p)$ and $f(n)=0$ for even $n$ . In particular, $S=f(2015!)=\boxed{0}$

Note that this function is a multiplicative function .

Note that each prime factor $\le 1007$ is being repeated at least twice, and so is 2. For example, the power of 2 in 2015! is 2005, so we can write the summation as: $\sum_{d|\color{#3D99F6}{2^{2005}}\cdot \color{#20A900}{\frac{2015!}{2^{2005}}}}\mu(d)\phi(d)=\color{#3D99F6}{\left(\sum_{d|2^{2005}} \mu(d)\phi(d)\right)}\color{#20A900}{\left(\sum_{d\mid\frac{2015!}{2^{2005}}} \mu(d)\phi(d)\right)}.$

The stuff in blue becomes $\color{#3D99F6}{\left(\sum_{d|2^{2005}} \mu(d)\phi(d)\right)}=\mu(1)\phi(1)+\mu(2)\phi(2)+\mu(2^2)\phi(2^2)+...+\mu(2^{2005})\phi(2^{2005})$

Then, we recall how the mobius function behaves over prime powers: $\mu(p^k)=\begin{cases}=&1&,&k=0\\=&-1&,&k=1\\=&0&,&k>1\end{cases}$ so, using this: $\mu(1)\phi(1)+\mu(2)\phi(2)+\mu(2^2)\phi(2^2)+...+\mu(2^{2005})\phi(2^{2005})=(1)(1)+(-1)(1)+0+0+0+...+0=0.$ By the zero product property , one part of the product is zero, so the product is $\boxed{0}$ .

Lets call $d_1$ is the divisors of 2015! which is square free. By definition of $\mu(d)$ we could easily see that $S=\sum_{d|2015!}\mu(d)\phi(d) =\sum_{d_1|2015!}\mu(d_1)\phi(d_1)$ $d_1$ could only take the form of $1, p_1,p_2,...,p_k, p_1*p_2,p_1*p_3,...,p_{k-1}*p_k, p_1*p_2*p_3,..., p_1*p_2*...*p_k$ , in which $p_k$ are the prime numbers less than 2015

So our sum will become $1 - (\phi(p_1)+\phi(p_2)+...+\phi(p_k)) +(\phi(p_1*p_2)+\phi(p_1*p_3)+...+\phi(p_{k-1}*p_k)) - (\phi(p_1*p_2*p_3)+...) +... -...$ This could be easily compact to $(1-\phi(p_1))(1-\phi(p_2))...(1-\phi(p_k))$

And note that $p_1=2$ , so $1-\phi(p_1)=\boxed{0}$