6. The set $X=\left\{1,2,3, \cdots, 5 \times 10^{6}\right\}$ has how many 2015-element subsets $A$ such that the sum of the elements of any non-empty subset of $A$ is not a multiple of 2016?
6. Let the set $A=\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\}$ satisfy the conditions of the problem.

Then $a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \cdots, a_{1}+a_{2}+a_{3}+\cdots+a_{2015}$ have distinct remainders when divided by 2016, and none of the remainders are 0.

Similarly, $a_{2}, a_{2}+a_{1}, a_{2}+a_{1}+a_{3}, \cdots, a_{2}+a_{1}+a_{3}+\cdots+a_{2015}$ have distinct remainders when divided by 2016, and none of the remainders are 0.
Thus, $a_{1} \equiv a_{2}(\bmod 2016)$.
Similarly,
$$
\begin{array}{l}
a_{1} \equiv a_{2}(\bmod 2016) \equiv \cdots \\
\equiv a_{2015}(\bmod 2016) \equiv k(\bmod 2016) .
\end{array}
$$

By the condition that the sum of elements of any non-empty subset of $A$ is not a multiple of 2016, we get $(k, 2016)=1$.
Note that, $2016=32 \times 9 \times 7$,
$$
5 \times 10^{6}=2016 \times 2480+320 \text {, }
$$

Euler's totient function $\varphi(2016)=16 \times 6 \times 6=576$.
Using the principle of inclusion-exclusion, the number of elements in $\{1,2, \cdots, 320\}$ that are coprime with 2016 is
$$
\begin{array}{l}
320-\left(\left[\frac{320}{2}\right]+\left[\frac{320}{3}\right]+\left[\frac{320}{7}\right]\right)+ \\
\left(\left[\frac{320}{2 \times 3}\right]+\left[\frac{320}{3 \times 7}\right]+\left[\frac{320}{2 \times 7}\right]\right)- \\
{\left[\frac{320}{2 \times 3 \times 7}\right]} \\
=92,
\end{array}
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$. Therefore, the number of subsets satisfying the conditions of the problem is
$$
92 \mathrm{C}_{2481}^{2015}+484 \mathrm{C}_{2488}^{2015} \text {. }
$$