Let $a_1 = \sqrt 7$ and $b_i = \lfloor a_i \rfloor$, $a_{i+1} = \dfrac{1}{b_i - \lfloor b_i \rfloor}$ for each $i\geq i$. What is the smallest integer $n$ greater than $2004$ such that $b_n$ is divisible by $4$? ($\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$)

$ 
\textbf{(A)}\ 2005 
\qquad\textbf{(B)}\ 2006
\qquad\textbf{(C)}\ 2007
\qquad\textbf{(D)}\ 2008
\qquad\textbf{(E)}\ \text{None of above}
$
1. We start with \( a_1 = \sqrt{7} \).
2. Calculate \( b_1 = \lfloor a_1 \rfloor = \lfloor \sqrt{7} \rfloor = 2 \).
3. Next, we compute \( a_2 \):
   \[
   a_2 = \frac{1}{a_1 - b_1} = \frac{1}{\sqrt{7} - 2}
   \]
   Rationalize the denominator:
   \[
   a_2 = \frac{1}{\sqrt{7} - 2} \cdot \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}
   \]
4. Calculate \( b_2 = \lfloor a_2 \rfloor = \lfloor \frac{\sqrt{7} + 2}{3} \rfloor = 1 \).
5. Next, we compute \( a_3 \):
   \[
   a_3 = \frac{1}{a_2 - b_2} = \frac{1}{\frac{\sqrt{7} + 2}{3} - 1} = \frac{1}{\frac{\sqrt{7} + 2 - 3}{3}} = \frac{3}{\sqrt{7} - 1}
   \]
   Rationalize the denominator:
   \[
   a_3 = \frac{3}{\sqrt{7} - 1} \cdot \frac{\sqrt{7} + 1}{\sqrt{7} + 1} = \frac{3(\sqrt{7} + 1)}{7 - 1} = \frac{3(\sqrt{7} + 1)}{6} = \frac{\sqrt{7} + 1}{2}
   \]
6. Calculate \( b_3 = \lfloor a_3 \rfloor = \lfloor \frac{\sqrt{7} + 1}{2} \rfloor = 1 \).
7. Next, we compute \( a_4 \):
   \[
   a_4 = \frac{1}{a_3 - b_3} = \frac{1}{\frac{\sqrt{7} + 1}{2} - 1} = \frac{1}{\frac{\sqrt{7} + 1 - 2}{2}} = \frac{2}{\sqrt{7} - 1}
   \]
   Rationalize the denominator:
   \[
   a_4 = \frac{2}{\sqrt{7} - 1} \cdot \frac{\sqrt{7} + 1}{\sqrt{7} + 1} = \frac{2(\sqrt{7} + 1)}{7 - 1} = \frac{2(\sqrt{7} + 1)}{6} = \frac{\sqrt{7} + 1}{3}
   \]
8. Calculate \( b_4 = \lfloor a_4 \rfloor = \lfloor \frac{\sqrt{7} + 1}{3} \rfloor = 1 \).
9. Next, we compute \( a_5 \):
   \[
   a_5 = \frac{1}{a_4 - b_4} = \frac{1}{\frac{\sqrt{7} + 1}{3} - 1} = \frac{1}{\frac{\sqrt{7} + 1 - 3}{3}} = \frac{3}{\sqrt{7} - 2}
   \]
   Rationalize the denominator:
   \[
   a_5 = \frac{3}{\sqrt{7} - 2} \cdot \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{3(\sqrt{7} + 2)}{7 - 4} = \frac{3(\sqrt{7} + 2)}{3} = \sqrt{7} + 2
   \]
10. Calculate \( b_5 = \lfloor a_5 \rfloor = \lfloor \sqrt{7} + 2 \rfloor = 4 \).

We observe that \( a_6 = a_2 \), indicating a periodic sequence with period 4. Therefore, \( b_5 = b_9 = b_{13} = \cdots \).

To find the smallest integer \( n \) greater than 2004 such that \( b_n \) is divisible by 4, we solve:
\[
n = 4k + 1 \quad \text{for} \quad k \in \mathbb{Z}
\]
\[
4k + 1 > 2004 \implies 4k > 2003 \implies k > 500.75 \implies k \geq 501
\]
Thus, the smallest \( n \) is:
\[
n = 4 \cdot 501 + 1 = 2005
\]

The final answer is \(\boxed{2005}\)