(1) According to the stated conditions we have $A_n(0, \frac{1}{n})$, and $B_n(b_n, \sqrt{2b_n})$ ($b_n > 0$). From $|OB_n| = \frac{1}{n}$ we get
$$b_n^2 + 2b_n = \left(\frac{1}{n}\right)^2.$$
Thus,
$$b_n = \sqrt{\left(\frac{1}{n}\right)^2 + 1} - 1, \quad n \in \mathbb{N}.$$

Since $2n^2b_n = 1 - n^2b_n^2 > 0$, we have $b_n + 2 = \frac{1}{n^2b_n}$ and
$$\begin{aligned} a_n &= \frac{b_n(1 + n\sqrt{2b_n})}{1 - 2n^2b_n} = \frac{b_n(1 + n\sqrt{2b_n})}{1 - (1 - n^2b_n^2)} \\ &= \frac{1}{n^2b_n} + \frac{\sqrt{2}}{\sqrt{n^2b_n}} = b_n + 2 + \sqrt{2(b_n + 2)}. \end{aligned}$$
Thus,
$$a_n = \sqrt{\left(\frac{1}{n}\right)^2 + 1} + 1 + \sqrt{2\sqrt{\left(\frac{1}{n}\right)^2 + 1} + 2}.$$
Since $\frac{1}{n} > \frac{1}{n+1} > 0$, $a_n > a_{n+1} > 4$ for any $n \in \mathbb{N}$.

(2) Let $c_n = 1 - \frac{b_{n+1}}{b_n}$ for $n \in \mathbb{N}$, then
$$\begin{align*} c_n &= \frac{\sqrt{\left(\frac{1}{n}\right)^2 + 1} - \sqrt{\left(\frac{1}{n+1}\right)^2 + 1}}{\sqrt{\left(\frac{1}{n}\right)^2 + 1} - 1} \\ &= n^2 \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right] \cdot \frac{\sqrt{\left(\frac{1}{n}\right)^2 + 1} + 1}{\sqrt{\left(\frac{1}{n}\right)^2 + 1} + \sqrt{\left(\frac{1}{n+1}\right)^2 + 1}} \\ &> \frac{2n+1}{(n+1)^2} \left[ \frac{1}{2} + \frac{1}{2\sqrt{\left(\frac{1}{n}\right)^2 + 1}} \right] > \frac{2n+1}{2(n+1)^2}. \end{align*}$$
Since $(2n+1)(n+2) - 2(n+1)^2 = n > 0$, we obtain
$$c_n > \frac{1}{n+2}, \quad n \in \mathbb{N}.$$
Let $S_n = c_1 + c_2 + \dots + c_n, \quad n \in \mathbb{N}$. If $n = 2^k - 2 > 1$ ($k \in \mathbb{N}$), then
$$\begin{align*} S_n &> \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^k - 1} + \frac{1}{2^k} \\ &= \left(\frac{1}{3} + \frac{1}{4}\right) + \left[\frac{1}{2^2 + 1} + \dots + \frac{1}{2^3}\right] + \dots + \left[\frac{1}{2^{k-1} + 1} + \dots + \frac{1}{2^k}\right] \\ &> 2 \cdot \frac{1}{2^2} + 2^2 \cdot \frac{1}{2^3} + \dots + 2^{k-1} \cdot \frac{1}{2^k} = \frac{k-1}{2}. \end{align*}$$
Therefore, if we put $n_0 = 2^{4009} - 2$, then for any $n > n_0$ we have
$$\begin{aligned} \left[1 - \frac{b_2}{b_1}\right] + \left[1 - \frac{b_3}{b_2}\right] + \dots + \left[1 - \frac{b_{n+1}}{b_n}\right] &= S_n > S_{n_0} > \frac{4009 - 1}{2} \\ &= 2004. \end{aligned}$$
Consequently,
$$$$
\frac{b_2}{b_1} + \frac{b_3}{b_2} + \cdots + \frac{b_n}{b_{n-1}} + \frac{b_{n+1}}{b_n} < n - 2004, \quad n > n_0.