62nd Ukrainian National Mathematical Olympiad

Without loss of generality, let $a_{101}$ be the largest of these numbers (or one of the largest). It is clear that $a_i+1\ge a_{i+1}$ for any $i=1,100$. If we add all these 100 inequalities, we get $a_1+a_2+...+a_{100}+100\ge a_2+a_3+...+a_{101}$, that is $a_1\ge a_{101}-100$. From the condition $a_{101}+1$ is divisible by $a_1$, and from the fact that $a_{101}$ is the largest we have $a_{101}+1>a_1$, so $a_{101}+1\ge 2a_1\ge 2(a_{101}-100)=2a_{101}-200$, therefore $a_{101}\le 201$. It remains to show that there is an example where the maximum is equal to 201. Indeed, consider the set 101, 102, ..., 201, for it $a_i+1=a_{i+1}$ for $i=1,100$ and $a_{101}+1=2a_1$, so this example satisfies the problem.