2. 57 A quirky mathematician has a ladder with $n$ steps, and he climbs up and down the ladder, each time ascending $a$ steps or descending $b$ steps. Here, $a, b$ are fixed positive integers. If he can start from the ground, climb to the very top of the ladder, and then return to the ground, find the minimum value of $n$ (expressed in terms of $a, b$) and prove it.
[Solution] The minimum value is $a+b-(a, b)$, where $(a, b)$ denotes the greatest common divisor of $a$ and $b$.

Since the level the mathematician reaches each time is a multiple of $(a, b)$, we can ignore other levels, treating $(a, b)$ as the first level, and using $\frac{a}{(a, b)}, \frac{b}{(a, b)}$ to replace $a$, $b$. Therefore, we only need to consider the case where $(a, b)=1$.
(1) $n=a+b-1$.
In this case, since $(a, b)=1$, then $a, 2a, 3a, (b-1)a, ba (\bmod b)$ are distinct residues, forming a complete residue system modulo $b$.

If the mathematician is at level $r$, and $r+a>n$, then it must be that $r>b$. The mathematician can descend several times until reaching level $n$. Thus, he is always confined to this residue class and cannot continue to walk to the other residue classes (especially the $ba$ class), which is a contradiction!
In summary, when $(a, b)=1$, the minimum value is $n=a+b-1$.