10. (40 points) On the board, there are 2015 pairwise distinct positive real numbers. It turns out that for any number $a>0$, the number of numbers on the board that are less than 2014/a, and the number of numbers that are greater than a, have the same parity. What can the product of all the numbers be?
Answer: $2014^{1007} \sqrt{2014}$.

Solution: We will prove that all numbers written on the board, except one, can be paired such that the product in each pair is 2014. Consider $a=\sqrt{2014}$. Since the number of numbers greater than it and the number of numbers less than it have the same parity, the number of numbers equal to it is odd, because 2015 is an odd number. Therefore, there is at least one number equal to $\sqrt{2014}$ in our set, and since all numbers are distinct, there is exactly one. Consider an arbitrary number $b$ and together with it the number $c=2014 / b$. Then, by the condition, the number of numbers less than $b$ and greater than $c$ have the same parity, and the number of numbers greater than $b$ and less than $c$ have the same parity. Thus, the number of numbers not equal to $b$ (greater + lesser) has the same parity as the number of numbers not equal to $c$. Since there are 2015 numbers in total, the number of numbers equal to $b$ and the number of numbers equal to $c$ have the same parity (here we take into account the fact that the set must contain the number $\sqrt{2014}$). Thus, if the number $b$ is in the set, then the number $2014 / b$ is also in the set.

Therefore, we have proved that all numbers can be paired with a product of 2014, except one equal to $\sqrt{2014}$, from which we easily obtain the answer.