7. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1} a_{n}-1=a_{n}^{2}$.
(1) Prove: $\sqrt{2 n-1} \leqslant a_{n} \leqslant \sqrt{3 n-2}$;
(2) Find the integer $m$, such that $\left|a_{2005}-m\right|$ is minimized. (2005 Hebei Province High School Mathematics Competition Problem)
7. (1) From the given information, it is easy to see that $\{a_n\}$ is an increasing sequence, and all its terms are positive.

Since $a_{n+1} a_{n} - 1 = a_{n}^2$, we have, for $k \geq 2$, $a_{k} = a_{k-1} + \frac{1}{a_{k-1}}$.
Thus, $a_{k}^2 = \left(a_{k-1} + \frac{1}{a_{k-1}}\right)^2 = a_{k-1}^2 + \frac{1}{a_{k-1}^2} + 2 > a_{k-1}^2 + 2$, which means $a_{k}^2 - a_{k-1}^2 > 2$.
Therefore, $a_{n}^2 - a_{1}^2 = \sum_{k=2}^{n} (a_{k}^2 - a_{k-1}^2) > 2(n-1)$, i.e., $a_{n}^2 > a_{1}^2 + 2(n-1) = 2n - 1$. Hence, $a_{n} > \sqrt{2n-1} (n > 1)$.
$$\text{Also, } a_{n}^2 - a_{1}^2 = \sum_{k=2}^{n} (a_{k}^2 - a_{k-1}^2) = 2(n-1) + \sum_{k=2}^{n} \frac{1}{a_{k-1}^2} \leq 2(n-1) + (n-1) = 3n - 3,$$ i.e., $a_{n}^2 \leq 3n - 3 + a_{1}^2 = 3n - 2$.

Thus, $a_{n} \leq \sqrt{3n-2}$. Combining these results, we have $\sqrt{2n-1} \leq a_{n} \leq \sqrt{3n-2}$.
(2) From the results in (1), we have $\sqrt{4009} \leq a_{2005} \leq \sqrt{6013}$, i.e., $631$ when $b_{n} > 0$.

The equation $a_{n+1}^2 = a_{n}^2 + \frac{1}{a_{n}^2} + 2$ can be rewritten as $2n + 1 + b_{n+1} = 2n - 1 + b_{n} + 2 + \frac{1}{2n - 1 + b_{n}}$.

This gives us
$$b_{n+1} = b_{n} + \frac{1}{2n - 1 + b_{n}} \leq b_{n} + \frac{1}{2n - 1}$$

By induction, we get
$$b_{n+1} \leq b_{1} + 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n-3} + \frac{1}{2n-1}$$

Since $b_{1} = 0$, we have $b_{2005} \leq 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{4005} + \frac{1}{4007}$.
To estimate the value on the right side, we segment it as follows:
$$\begin{aligned}
b_{2005} \leq & 1 + \left(\frac{1}{3} + \frac{1}{5} + \frac{1}{7}\right) + \left(\frac{1}{9} + \frac{1}{11} + \cdots + \frac{1}{25}\right) + \cdots + \left(\frac{1}{243} + \frac{1}{245} + \cdots + \frac{1}{727}\right) + \\
& \left(\frac{1}{729} + \frac{1}{731} + \cdots + \frac{1}{2185}\right) + \\
& \left(\frac{1}{2187} + \cdots + \frac{1}{4005} + \frac{1}{4007}\right)
\end{aligned}$$

The first parenthesis on the right side contains 3 terms, with the largest term being $\frac{1}{3}$; the second parenthesis contains 9 terms, with the largest term being $\frac{1}{9}$; ...; the sixth parenthesis contains 729 terms, with the largest term being $\frac{1}{729}$; finally, the seventh parenthesis contains 911 terms, with the largest term being $\frac{1}{2187}$. Therefore, $b_{2005} < 8$. Combining $a_{n}^2 = 2n - 1 + b_{n}$, we get $a_{2005}^2 < 4010 - 1 + 8 < 4032.25 = 63.5^2$.

Thus, $63 < a_{2005} < 63.5$. Hence, $m = 63$.