Observe that $d(1008) + d(1009) + \cdots + d(2015) = d(1)+d(2) + \cdots + d(2015) - (d(1)+d(2) \cdots + d(1007)) = d(1) + d(2) + \cdots + d(2015) - (d(2)+d(4) + \cdots + d(2014)) = d(1)+d(3) +\cdots + d(2015) = 1008^2$

One can show in general that if $$f(n)=d(n)+d(n+1)+d(n+2)+...+d(2n-1)$$ then $$f(n)=n^2$$ for all $n\in\mathbb{N}$. Indeed, $f(1)=d(1)=1$, and $$f(n+1)=f(n)-d(n)+d(2n)+d(2n+1)=f(n)-d(n)+d(n)+2n+1=f(n)+2n+1$$ and then the claim follows by induction.

More simply, $S = \{d(n), .., d(2n-1)\}$. Notice that since $S \subseteq \{1, 3, ..., 2n-1\}$, $\mid S \mid = n$, and $d(i) \neq d(j)$ for $i \neq j \Longrightarrow S = \{1, 3, ..., 2n-1\}$, from which the conclusion follows.