Positive integers $a_1, a_2, \ldots, a_{101}$ are such that $a_i+1$ is divisible by $a_{i+1}$ for all $1 \le i \le 101$, where $a_{102} = a_1$. What is the largest possible value of $\max(a_1, a_2, \ldots, a_{101})$?

[i]Proposed by Oleksiy Masalitin[/i]
1. **Assume the largest value:**
   Let \( a_{101} \) be the largest of the \( a_i \), and denote \( a_{101} = k \). We aim to find the maximum possible value of \( k \).

2. **Claim:**
   We claim that \( a_{101-i} = k - i \) for all \( 0 \leq i \leq 100 \).

3. **Base case:**
   For \( i = 0 \), we have \( a_{101-0} = a_{101} = k \), which is trivially true.

4. **Inductive step:**
   Assume \( a_{101-i} = k - i \) for some \( i \). We need to show that \( a_{101-(i+1)} = k - (i+1) \).

5. **Divisibility condition:**
   Since \( a_{101-i} = k - i \), we have \( k - i \mid a_{101-(i+1)} + 1 \).

6. **Bounding \( a_{101-(i+1)} \):**
   If \( a_{101-(i+1)} + 1 \geq 2(k - i) \), then:
   \[
   k + 1 \geq a_{101-(i+1)} + 1 \geq 2(k - i)
   \]
   Simplifying, we get:
   \[
   k + 1 \geq 2k - 2i \implies k + 1 \geq k + k - 2i \implies 1 \geq k - 2i
   \]
   Since \( k > 201 \) and \( i \leq 100 \), this inequality cannot hold. Therefore, \( a_{101-(i+1)} + 1 < 2(k - i) \).

7. **Conclusion from the bound:**
   Thus, \( a_{101-(i+1)} + 1 = k - i \), which implies \( a_{101-(i+1)} = k - (i+1) \).

8. **Inductive proof:**
   By induction, \( a_{101-i} = k - i \) for all \( 0 \leq i \leq 100 \).

9. **Relating \( a_1 \) and \( a_{101} \):**
   From the claim, \( a_1 = k - 100 \). Since \( a_1 \mid a_{101} + 1 \), we have:
   \[
   k - 100 \mid k + 1
   \]
   This implies:
   \[
   k - 100 \mid 101
   \]

10. **Divisors of 101:**
    The divisors of 101 are 1 and 101. Since \( k - 100 > 101 \), the only possible value is \( k - 100 = 101 \), giving \( k = 201 \).

11. **Verification:**
    If \( k = 201 \), then \( a_1 = 101 \), \( a_2 = 102 \), ..., \( a_{101} = 201 \). Each \( a_i + 1 \) is divisible by \( a_{i+1} \), satisfying the problem's conditions.

The final answer is \( \boxed{201} \).