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.