1. Define the Primes:
Let $p_1, p_2, \ldots, p_{2009}$ be the first 2009 prime numbers. Let $q_1, q_2, \ldots, q_{2009}$ be another set of prime numbers such that $q_i > 20p_i$ for all $1 \leq i \leq 2009$. This is possible because there are infinitely many primes, and for any prime $p_i$, there exists a prime $q_i$ such that $q_i > 20p_i$.

2. Chinese Remainder Theorem:
By the Chinese Remainder Theorem, there exists an integer $a$ such that:
$$a + i \equiv 0 \pmod{p_i q_i} \quad \text{for all} \quad 1 \leq i \leq 2009$$
This is possible because the moduli $p_1 q_1, p_2 q_2, \ldots, p_{2009} q_{2009}$ are pairwise coprime.

3. Consecutive Integers:
Consider the sequence $a+1, a+2, \ldots, a+2009$. For each $i$ in this range, $a+i$ is divisible by $p_i q_i$. Therefore, the smallest prime divisor of $a+i$ is $p_i$ and the largest prime divisor is $q_i$.

4. Ratio of Prime Divisors:
For each $a+i$, the ratio between the largest and the smallest prime divisor is:
$$\frac{q_i}{p_i} > 20$$
This satisfies the condition that the ratio between the largest and the smallest prime divisor is more than 20.

5. Conclusion:
Thus, the sequence $a+1, a+2, \ldots, a+2009$ consists of 2009 consecutive positive integers such that for each of them, the ratio between the largest and the smallest prime divisor is more than 20.

$\blacksquare$