7・84 Let $a$ and $b$ be given positive integers. There is a robot moving up and down a staircase with $n$ steps. The robot ascends exactly $a$ steps each time it goes up and descends exactly $b$ steps each time it goes down. To ensure that the robot can reach the top of the staircase from the ground and then return to the ground after several steps of ascending and descending, what is the minimum value of $n$? Prove your conclusion.
[Solution] We refer to the ground as the 0th level of the staircase and count the levels from bottom to top. When we invert the directions of up and down, $a$ and $b$ are swapped, so without loss of generality, assume $a \geqslant b$.

If $b \mid a$, then we can set $a=s b, s \in \mathbb{N}$. Taking $n=a$, the robot ascends once and then descends $s$ times, thus moving from the ground to the top of the staircase and then returning to the ground. Clearly, for the case of $n1$, we can set $a=a_{1} d, b=b_{1} d$, so $\left(a_{1}, b_{1}\right)=1$. Applying the same discussion to $a_{1}$ and $b_{1}$, we find that the minimum value of $n$ in this case is
$$
d\left(a_{1}+b_{1}-1\right)=a+b-(a, b) .
$$

Clearly, this result is consistent with the results of the previous two cases, so the minimum value of $n$ sought is $a+b-(a, b)$.