Answer is Anja wins if $n$ is odd and if $n$ is not a power of $2$ and Bernd wins if $n$ is a power of 2. See below for the proof.

The above "solution" gives the correct result but not with a proper reasoning: You cannot speak of $n$ being a winning or lost position because it is also "part of the position" how many stones you are allowed to take in your next move. So a priori it might well happen that a situation with $n$ stones and no restriction on the number of stones to take in your next move is winning, but if you are only allowed to take a certain number it is losing. However, we can argue differently: As was mentioned above, once it's your move with an odd number of stones, you are winning because you can just take one stone and force the rest of the game to consist of one-stone-moves. (Note that here it doesn't matter how the number of stones to take is restricted before your move because you are always allowed to take just one stone). This means that we only have to consider the case where $n$ is even and we may also suppose that in each turn, the player takes an even number of stones since otherwise he will be losing. But then we can just divide all the numbers by $2$ and continue as before! From here it is immediate that any power of $2$ is losing but every other $n$ is winning.

XbenX wrote: If $n$ is even , let $n=2k$ , if $k$ is odd then Anja takes $2$ wich forces Bernd to take $2$ and with the same reasoning as above Anja wins. If $n$ is a power of $2$ then Anja must take a number less than the half , so the number of stones left will be $2k$ for $k$ odd , and this turns to the above case, with Bernd starting , in this case Bernd wins. Now you removed your previous mistake. What you wrote is correct, but you don't consider all the cases. For instance, $n=12$ is not covered. Of course this can be modified again but I guess you will end up with my solution in that case.