First let us change the order: $a_1 = b_m$, $a_2 = b_{m-1}$, ..., $a_m = b_1$. Then we have $1 \le b_m \le b_{m-1} \le \dots \le b_2 \le b_1 \le 400$. Let us show that $b_k \le \frac{400}{k}$, $1 \le k \le m$. If $k=1$, it is true.

For the induction step, we need to prove that $b_{k+1} \le \frac{400}{k+1}$. By the condition
$$b_k - b_{k+1} \ge (b_k, b_{k+1}) = \frac{b_k \cdot b_{k+1}}{[b_k, b_{k+1}]} \ge \frac{b_k \cdot b_{k+1}}{400}$$
holds. From this inequality we have $b_{k+1} \le \frac{400b_k}{b_k + 400} = \frac{400}{1 + \frac{400}{b_k}} \le \frac{400}{1+k}$. Hence $\frac{400}{b_k} \ge k$.

The $1 \le b_m \le b_{m-1} \le \dots \le b_{20} \le \frac{400}{20} = 20$. Since $b_i$ is integer for $1 \le i \le m$, $m \le 40$.