We start by proving that it is always possible either to make all of the coins face up or to make all of the coins face down.
Notice that if we proceed by choosing the numbers $i=1,2, \ldots, 2015$ in this order or in any other order, the final result will depend only on the $i$ coins chosen at each step. For the sake of simplicity, we will follow the order
$$i=1007,1008,1006,1009,1005,1010, \ldots, 2,2013,1,2014,2015 .$$
By symmetry of the face up/face down positions, we can assume without loss of generality that at the beginning, the number of coins face up is odd equal to $2k+1$. Choose $k$ of these face up coins and $1007-k$ of the face down coins and turn them over to get $1008$ coins face up and $1007$ coins face down. Turn these new $1008$ coins face up to get all the coins face down. After this, at each step, whenever we choose $i$ coins, for $i=1006,1005, \ldots, 1,2015$, in the next step we choose the $2015-i$ remaining coins to make at the end all the coins face up.

Now we prove that for each starting position, it is not possible to choose to make all the coins face up or all the coins face down.
Assume that starting with an odd number of face up coins we can make at the end all the coins face down. This means that each face up coin will be turned over an odd number of times while each face down coin will be turned over an even number of times. Therefore, the total number of turning over coins is
$$\text{odd} \times \text{odd} + \text{even} \times \text{even} = \text{odd} .$$
But the total number of turning over coins is
$$1+2+\cdots+2015=2015 \times 1008=\text{even},$$
which is a contradiction. So it is only possible to make all the coins face up and not to make all of them face down. In a similar way, if we start with an odd number of face down coins it is only possible to make them all face down.

The statement works for any odd number of coins. We prove it by induction on the number $n$ of coins, $n$ being odd. The case $n=1$ is obvious. Suppose the statement is true for $n=2k-1$, for some positive integer $k$. If we are given $n=2k+1$ coins, we consider the following cases.

- Case 1. There is a coin $C_{1}$ facing up and another coin $C_{2}$ facing down. We consider the other $2k-1$ coins first. By our induction, we can turn the coins $1,2, \ldots, 2k-1$ times in succession so all the $2k-1$ coins are in one direction. Without loss of generality, we assume that all the $2k-1$ coins are facing up. Then we turn all these coins together with $C_{1}$ and then turn all the $2k+1$ coins so they will be all facing up.

- Case 2: All the coins are in the same direction. We arrange the coins around a circle and number them $1,2, \ldots, 2k+1$ in clockwise order. We first turn coin $1$, then coins $2$ and $3$, and then $4,5$, and $6$, and so on along the circle. Then we make a total of $1+2+\cdots+(2k+1)=(k+1)(2k+1)$ turnings and each coin has been turned $k+1$ times. Since they start in the same direction, they end in the same direction.

From the above argument, we can find a way to make all of the $2k+1$ coins facing in one direction after $2k+1$ operations regardless of the initial configuration. Hence our induction is complete.

To prove that it is impossible to achieve both final configurations, we proceed as in the first solution.