9. Given $f(x)=\left\{\begin{array}{l}\frac{3+x}{1+x^{2}}, 0 \leq x \leq 3, \\ \frac{3}{5}, x>3 .\end{array}\right.$, the sequence $\left\{a_{n}\right\}$ satisfies: $0<a_{n} \leq 3, n \in N^{*}$, and $\sum_{i=1}^{2009} a_{i}=\frac{2009}{3}$, if the inequality $\sum_{i=1}^{2009} f\left(a_{i}\right) \leq x-\ln (x-p)$ holds for all $x \in(p,+\infty)$, find the minimum value of the real number $p$.
9. Analysis: Since $a_{1}+a_{2}+a_{3}+\square+a_{2009}=\frac{2009}{2}$, when $a_{1}=a_{2}=a_{3}=\square=a_{2009}=\frac{1}{3}$,

we have $f\left(\frac{1}{3}\right)=3$, so $f\left(a_{1}\right)+f\left(a_{2}\right)+f\left(a_{3}\right)+\square+f\left(a_{2009}\right)=2009 f\left(\frac{1}{3}\right)=6027$ holds.
First, we prove: $f\left(a_{1}\right)+f\left(a_{2}\right)+f\left(a_{3}\right)+\square+f\left(a_{2009}\right) \leq 6027$.
It is easy to know that $f(x)=\frac{3+x}{1+x^{2}}(0 \leq x \leq 3)$ has the tangent line equation at $x=\frac{1}{3}$ as $y=\frac{3}{10}(11-3 x)$.
Thus, $f(x)=\frac{3+x}{1+x^{2}} \leq \frac{3}{10}(11-3 x) \Leftrightarrow(x-3)\left(x-\frac{1}{3}\right)^{2} \leq 0$ holds.
So when $0p$. Let $g^{\prime}(x)=0$, we get $x=p+1$,
when $pp+1$, $g^{\prime}(x)>0$, at this time $g(x)$ is monotonically increasing.
Therefore, $g(x)_{\min }=g(p+1)=p+1$
To make the inequality $f\left(a_{1}\right)+f\left(a_{2}\right)+f\left(a_{3}\right)+\square \quad f\left(a_{2009}\right) \leq x-\ln (x-p)$ hold for all $x \in(p,+\infty)$, we only need $6027 \leq p+1$, which gives $p \geq 6026$, i.e., the minimum value of $p$ is 6026.