18. (15 points) Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and its first $n$ terms sum $S_{n}$ satisfies $S_{n}=\frac{1}{4}\left(a_{n}-1\right)\left(a_{n}+3\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{S_{n}}$, try to find the first $n$ terms sum $T_{n}$ of $\left\{b_{n}\right\}$.
18. (1) From $a_{1}=S_{1}=\frac{1}{4}\left(a_{1}-1\right)\left(a_{1}+3\right)$ and $a_{n}>0$, we get $a_{1}=3$.

From $S_{n}=\frac{1}{4}\left(a_{n}-1\right)\left(a_{n}+3\right)$, we get $S_{n-1}=\frac{1}{4}\left(a_{n-1}-1\right)\left(a_{n-1}+3\right)$.
Thus, $a_{n}=\frac{1}{4}\left[\left(a_{n}^{2}-a_{n-1}^{2}\right)+2\left(a_{n}-a_{n-1}\right)\right]$.
Rearranging gives $2\left(a_{n}+a_{n-1}\right)=\left(a_{n}+a_{n-1}\right)\left(a_{n}-a_{n-1}\right)$.
Since $a_{n}+a_{n-1}>0$, we have $a_{n}-a_{n-1}=2$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 3 and common difference 2.
Hence, $a_{n}=2 n+1$.
(2) Since $a_{n}=2 n+1$, we have,
$$
\begin{array}{l}
S_{n}=n(n+2), b_{n}=\frac{1}{S_{n}}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right), \\
T_{n}=b_{1}+b_{2}+\cdots+b_{n} \\
=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+2}\right)\right] \\
=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right) \\
=\frac{3}{4}-\frac{2 n+3}{2(n+1)(n+2)} .
\end{array}
$$