Check that $A_k$ has $2k-1$ elements and the largest element of $A_k$ is $k^2$. Consequently, the smallest element of $A_k$ is $k^2-(2k-1)+1=k^2-2k+2$. Thus, $b_k -1=k(k-1)$. We then compute \[ \sum_{2\le i\le 2000}\frac{2000}{b_i-1} = 2000\sum_{2\le i\le 2000}\frac{1}{i(i-1)} = 2000\sum_{2\le i\le 2000}\left(\frac{1}{i-1}-\frac{1}{i}\right)=1999. \]Note that it should start from $i=2$ as $b_1-1=0$.