Austrian Mathematical Olympiad

Answer. Mr. Precise can ensure his victory when $n$ is a power of 2.

We label the positions consecutively $0$, $1$, ..., $n-1$ where $0$ is the front position. If $n$ is a power of $2$, say $n=2^k$, Mr. Precise can simply always put in the current position as number of seconds. If the microwave turns the plate backwards, the tea cup will end up front immediately. Otherwise, the number of the position will be doubled and reduced modulo $2^k$ at each turn and therefore be divisible by $2^k$ after at most $k$ turns. This means that the tea cup ends up front.

Now, let $n=2^k\cdot m$, where $m>1$ is an odd number. We will show that the microwave can always choose a number not divisible by $m$. This is clearly true for the first position, for example by choosing the position $1$. After that, the tea cup is in a certain position $p$ not divisible by $m$ and Mr. Precise puts in $s$ seconds. If both $p+s$ and $p-s$ were divisible by $m$, this would also be true for the sum, so that $m\mid 2p$. Since $m$ is odd, this implies $m\mid p$ which is wrong.