9. (16 points) Let $f(x)=x+\frac{1}{x}(x>0)$. If for any positive number $a$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in the interval $\left[1, a+\frac{2013}{a}\right]$, such that the inequality
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right)<f\left(a_{m+1}\right)
$$
holds, find the maximum value of $m$.
II. 9. Let $a=\sqrt{2013}$, then there exist $m+1$ real numbers that meet the requirements in the interval $[1,2 \sqrt{2013}]$.
Notice that $[1,2 \sqrt{2013}] \subseteq\left[1, a+\frac{2013}{a}\right]$.
Therefore, we only need to consider the existence of real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in $[1,2 \sqrt{2013}]$.

It is easy to see that $f(x)$ is an increasing function in the interval $[1,2 \sqrt{2013}]$.
Thus, $f(1) \leqslant f\left(a_{i}\right)(i=1,2, \cdots, m)$, $f\left(a_{m+1}\right) \leqslant f(2 \sqrt{2013})$.
Adding the first $m$ inequalities, we get
$$
\begin{array}{l}
m f(1) \leqslant f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right) \\
<f\left(a_{m+1}\right) \leqslant f(2 \sqrt{2013}),
\end{array}
$$

which implies $m<\sqrt{2013}+\frac{1}{4 \sqrt{2013}}<45$.
Therefore, $m \leqslant 44$.
When $m=44$, take $a_{1}=a_{2}=\cdots=a_{44}=1$, $a_{45}=2 \sqrt{2013}$, then the inequality in the problem holds.
Hence, the maximum value of $m$ is 44.