We claim that no couples remain at the end, giving an answer of $\boxed 0$. Relabel the $0$ as a $2018$ for convenience. Define a "tick" to be $2018$ iterations of minutes (i.e. the $1$st-$2018$th minutes, $2019$th-$4036$th minutes, etc.), so that there are $2018$ ticks in total. Note that each couple moves every tick, so if a couple remains at the end, it will have moved $2018$ times. In particular, it will have visited $2019$ points. On the other hand, since it is on an even point after time $2018$, we know that it will have visited a repeat point by $1011$ ticks. We claim that this is impossible. Indeed, suppose the couple does $a\to 2a$ on some move. Then, it moves again in $a$ minutes. On the other hand, after $1009$ minutes, if the couple has not left $2a$, then it will drop out. In particular, this means that it continues if and only if $a \leq 1009 \iff 2a \pmod{2018} > a$. Thus, if the couple visits point $p$ twice, and it does $p\to 2p \to \cdots \to 2^kp \to 2^{k+1}p = p$, then we cannot have $p < 2p < \cdots < 2^kp < p$ (taken modulo $2018$), so one of $2^ip \geq 2^{i+1}p$, i.e. the couple drops out after landing at $2^{i+1}p$. Since the couple visits $p$ by the end of $1011$ ticks, it will drop out before the end, since it will drop out within $1012$ ticks. $\blacksquare$ Edward Wan remarks that in fact the couple cannot do $p\to 2p\to \cdots \to 4096p$, since we cannot have $p < 2p < \cdots < 2048p$. So, the couple moves at most $11$ times before dropping out, i.e. the couples all drop out after at most $11+1$ ticks, or $2018\cdot 12$ minutes. The real bound is most certainly smaller. Remark 2: In fact, if we do $p \to \cdots \to 2^kp$, where $2^{k-1}p > 1009$ is minimal, then the couple arrives at $2^kp$ within the first tick. In the second tick, the couple will be sent away, and this will happen at or before the $3027$th minute, since $2^{k-1}p + 1009 \pmod{2018} \leq 1009$. Equality holds when the couple starts at $1009$, so the real bound is $3027$ minutes.

$(m+1)/2$, in which $n=2^\alpha m$, $m$ odd.

Write $n=2^\alpha m$ as in the hidden answer. Then, after $\alpha$ steps, you can verify that only the multiples of $2^\alpha$ remain. Ignore the operation at zero, as it does nothing. Now it remains to solve the problem for $m$ odd. But this is a matter of performing the operation and checking small cases: after $m$ steps, the remaining numbers are $0$ and the odd numbers from $1$ to $m-2$. Then, after $(m-1)/2$ steps the numbers are $0$ and all numbers from $(m+1)/2$ to $m-1$. If you apply the operation $(m-1)/2$ more times, they revert back to $0$ and the odd numbers, and we have a cycle, and we are basically done.

505