19. From the conditions, when $m \mid n$ and $m < n$, we have $a_{n} \geqslant 2 a_{m}$. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{4} \geqslant 2 a_{2} \geqslant 2^{2}$, similarly, $a_{8} \geqslant 2^{3}, a_{16} \geqslant 2^{4}, a_{30} \geqslant 2^{5}, a_{400} \geqslant 2^{6}, a_{2000} \geqslant 2^{7}$, which means $a_{2000} \geqslant 128$.

On the other hand, for any positive integer $n$, let the prime factorization of $n$ be
$$n=p_{1}^{q_{1}} p_{2}^{q_{2}} \cdots p_{k^{\prime}}^{q_{k}}$$

where $p_{1}<p_{2}<\cdots<p_{k}$ are prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are positive integers. Define
$$a_{n}=2^{a_{1}+a_{2}+\cdots+a_{k}}$$

Then the sequence $\left\{a_{n}\right\}$ satisfies the requirements of the problem, and
$$a_{2000}=2^{4+3} \leqslant 2^{7}$$

Therefore, the minimum value of $a_{2000}$ is 128.
19 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integers $m, n$, if $m \mid n, m<n$, then $a_{m} \mid a_{n}$, and $a_{m}<a_{n}$. Find the minimum possible value of $a_{2000}$.