9. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy:
$$
a_{1}=1, a_{n+1}=a_{n}+\frac{1}{a_{n}}(n=1,2, \cdots) \text {. }
$$

Find the integer part of $a_{2019}$.
From the given conditions,
$$
\begin{array}{l}
a_{n+1}^{2}=a_{n}^{2}+\frac{1}{a_{n}^{2}}+2 \\
\Rightarrow a_{n+1}^{2}-a_{n}^{2}>2 \\
\Rightarrow a_{n}^{2}>a_{1}^{2}+2(n-1)=2 n-1 \\
\Rightarrow a_{2019}^{2}>4037 .
\end{array}
$$

From $a_{n}^{2}=a_{n-1}^{2}+\frac{1}{a_{n-1}^{2}}+2$,
$$
\begin{array}{l}
a_{n-1}^{2}=a_{n-2}^{2}+\frac{1}{a_{n-2}^{2}}+2, \\
\cdots \cdots \\
a_{2}^{2}=a_{1}^{2}+\frac{1}{a_{1}^{2}}+2,
\end{array}
$$

Summing up, we get
$$
\begin{array}{l}
a_{n}^{2}=a_{1}^{2}+\sum_{i=1}^{n-1} \frac{1}{a_{i}^{2}}+2(n-1) \\
=2 n-1+\sum_{i=1}^{n-1} \frac{1}{a_{i}^{2}} \\
<2 n-1+\sum_{i=1}^{n-1} \frac{1}{2 i-1} .
\end{array}
$$

Thus, $a_{2019}^{2}$
$$
\begin{array}{l}
<4037+\left(1+\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}\right)+\right. \\
\left(\frac{1}{9}+\cdots+\frac{1}{15}\right)+\cdots+ \\
\left.\left(\frac{1}{2049}+\cdots+\frac{1}{4035}\right)\right) \\
<4037+\left(1+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\right. \\
\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\cdots+ \\
\left.\left(\frac{1}{2048}+\cdots+\frac{1}{2048}\right)\right) \\
<4044 .
\end{array}
$$

Therefore, $4037<a_{2019}^{2}<4044$, which means the integer part of $a_{2019}$ is 63.