1. $ka, kb$ is similar to $a, b$ after dividing by $k$, so we can assume $\gcd(a, b) = 1$. 2. The first machine never pauses working. 3. The moment from which both start working consecutively is the smallest multiple of $a$ which is $\equiv -1 \pmod b$. 4. If there have been $k$ operations of filling, at time $ka$ exactly $\left\lfloor \frac{(k-1)a}{b} \right\rfloor$ emptying operations started before this time.

The answer is $n=2a+2b-2\gcd(a,b).$ First, note that if $\gcd(a,b)\neq1,$ all values will be a multiple of that gcd, so we can scale it down and the answer will be $\gcd(a,b) \cdot \left(\frac{a}{\gcd(a,b)} + \frac{b}{\gcd(a,b)} - 2\right).$ Thus, it is sufficient to prove that the answer is $n=2a+2b-2$ for coprime integers $a$ and $b,$ so assume $\gcd(a,b)=1.$ Claim: $n = 2a + 2b - 2$ works once equilibrium is reached. Note that if both machines work without stopping, the state will cycle every $\text{lcm}(a,b)$ minutes, and within this time, there will be some point at which both machines simultaneously move to a new group of cups. Let us denote this time as time $0.$ Let $X$ be the number of empty cups plus the $a$ cups being filled at time $0 + \epsilon$ where $\epsilon$ is some extremely small number. Similarly, let $Y$ be the number of filled cups plus the $b$ cups being emptied at time $0 + \epsilon.$ NoteIt might be easier to think of it as where the machine pauses for a very, very small amount of time between switching groups when no cups are in a machine, and then assign $X$ as the number of empty cups and $Y$ as the number of filled cups at time $0.$ We will have $n = X + Y.$ At time $kb,$ the emptying machine will need to switch to another group of $b$ filled cups to start emptying. At time $0$ there were $Y$ filled cups. Since then, there have been $kb$ additional cups emptied but also $\left\lfloor \frac{kb}{a} \right\rfloor \cdot a$ cups filled. This gives us the inequality $$Y - kb + \left\lfloor \frac{kb}{a} \right\rfloor \cdot a \ge b \implies Y \ge b + (bk \text{ mod } a)$$which must be satisfied for all nonnegative integers $k.$ Since $a$ and $b$ are coprime, $bk \text{ mod } a$ can equal any residue from $0$ to $a-1$ for some $k.$ Thus, we must have $$Y \ge b + (a-1).$$By studying the other machine, we can also similarly get $$X \ge a + (b-1).$$Since $n = X + Y,$ we have $$n \ge 2(a + b - 1).$$However, once the machines have reached this equilibrium state, the inequalities above guarantee that both machines will never have a time when they cannot switch to a new group, which means they can infinitely. Claim: The equilibrium stated above will be reached for $n = 2a + 2b - 2.$ First, note that any period of time when both machines are running simultaneously will begin at a moment where both machines take new groups simultaneously (otherwise whichever machine started running after a break or for the first time could, and thus would, have started earlier. Now, using the logic above, we can prove by induction that the filling machine will never stop running. Assume that at time $ak$ the filling machine has not stopped running yet. There are initially $2a-2b-2$ empty cups. $ak$ of those will have been emptied, and less than or equal to $\left\lfloor \frac{ak}{b}\right\rfloor \cdot b$ have been emptied. We proved above that this number is sufficient that the machine will always be able to switch. Next, make the following observations: 1. At the initial moment of a period when both machines are simultaneously running, both machines will have moved to a new group of cups, as otherwise the machine which had not been running previously could have started earlier). 2. There are initially $n$ empty cups. In order for the filling machine to not stop running in a period of equilibrium, at the start of the period there must have been $X=n/2$ cups. Similarly, for the emptying machine to run, there must have been $Y=n/2$ cups. We know that $X$ always remains greater than $n/2$ from our earlier observation that the filling machine never stops running, so if $Y$ ever reaches greater than or equal to $n/2,$ it has reached a period of equilibrium when neither machine will stop running. 3. Finally, notice that both machines are running at the same rate of cups per minute. Thus, at the beginning of a period when both machines run at the same time if the filling machine has run $k$ times and the emptying machine has run $k'$ times, the number of filled cups (the value $Y$ from above) upon entering the period will be $$ka - k'b.$$Note that this value is exactly equal to the number of minutes when the filling machine runs while the emptying machine does not run. Since a period when both machines are simultaneously running which eventually ends due to one machine not being able to continue is capped at $\text{lcm}(a,b)-1$ minutes (otherwise it will have cycled and thus have entered equilibrium), there is a finite amount of time after which $Y$ will have had to have reached a value greater than or equal to $n/2,$ which from our observation above proves that it will reach equilibrium and neither machine will stop running. Thus, the answer is $\boxed{2a + 2b - 2\gcd(a,b)}.$ (Solved with input from yofro).