Since $a\geq 1$ and $$a_1\mid a_2\mid a_{10}\mid a_{20}\mid a_{100}\mid a_{200}\mid a_{1000}\mid a_{2000}$$ we have $a_{2000}\geq 2^7=128$, since they are strictly increasing. This is possible by considering the sequence $a_n=2^{\sum{b_i}}$, where $n=p_1^{b_1}p_2^{b_2}...p_k^{b_k}$ where $p_1,...p_k$ are all distinct primes.

$2000=2^4*5^3$. Therefore $a_{2000}=k_1a_{1000}=k_1*k_2*a_{500}=k_1*k_2*k_3*k_4*k_5*k_6*k_7*a_1$, were $k_i>1$ and integer. Least $a_{2000}=2^7$. We can define $a_n=2^{m_1+m_2+...+m_k}$ for $m=\prod_i p_i^{m_i}.$