Do there exist postive integers $a_1<a_2<\cdots<a_{100}$ such that for $2\le k\le100$ the greatest common divisor of  $a_{k-1}$ and $a_k$ is greater than the greatest common divisor of $a_k$ and $a_{k+1}$?
1. **Define the sequence**: We need to construct a sequence of positive integers \(a_1 < a_2 < \cdots < a_{100}\) such that for \(2 \le k \le 100\), the greatest common divisor (gcd) of \(a_{k-1}\) and \(a_k\) is greater than the gcd of \(a_k\) and \(a_{k+1}\).

2. **Prime number notation**: Let \(p_n\) denote the \(n\)-th prime number. We will use prime numbers to construct our sequence.

3. **Construct the sequence**:
   - For \(n \ge 3\), define:
     \[
     a_n = p_{101-n+1} \cdot p_{101-n} \cdot \prod_{i=1}^{n} p_{m+i}
     \]
     where \(m\) is an arbitrarily large positive integer (e.g., \(m = 500\)).
   - Define the first two terms as:
     \[
     a_1 = p_{101}, \quad a_2 = p_{101} \cdot p_{100}
     \]

4. **Verify the gcd conditions**:
   - For \(k \ge 3\), we need to check the gcd conditions:
     \[
     \gcd(a_{k-1}, a_k) = \gcd(p_{101-(k-1)+1} \cdot p_{101-(k-1)} \cdot \prod_{i=1}^{k-1} p_{m+i}, p_{101-k+1} \cdot p_{101-k} \cdot \prod_{i=1}^{k} p_{m+i})
     \]
     Since \(a_{k-1}\) and \(a_k\) share the prime factor \(p_{101-k+1}\), we have:
     \[
     \gcd(a_{k-1}, a_k) = p_{101-k+1}
     \]
   - Similarly, for \(k \ge 3\):
     \[
     \gcd(a_k, a_{k+1}) = \gcd(p_{101-k+1} \cdot p_{101-k} \cdot \prod_{i=1}^{k} p_{m+i}, p_{101-(k+1)+1} \cdot p_{101-(k+1)} \cdot \prod_{i=1}^{k+1} p_{m+i})
     \]
     Since \(a_k\) and \(a_{k+1}\) share the prime factor \(p_{101-k}\), we have:
     \[
     \gcd(a_k, a_{k+1}) = p_{101-k}
     \]

5. **Compare the gcd values**:
   - For \(2 \le k \le 100\):
     \[
     \gcd(a_{k-1}, a_k) = p_{101-k+1} > p_{101-k} = \gcd(a_k, a_{k+1})
     \]

6. **Verify the increasing sequence**:
   - By construction, \(a_n\) is defined such that each term is larger than the previous term due to the multiplication of additional prime factors. Thus, \(a_n > a_{n-1}\).

Therefore, the sequence \(a_1, a_2, \ldots, a_{100}\) satisfies the given conditions.

The final answer is \( \boxed{ a_1, a_2, \ldots, a_{100} } \) as defined above.