20. (12 points) Given the function $f(x)=\frac{2 x+3}{3 x}$, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n+1}=f\left(\frac{1}{a_{n}}\right)\left(n \in \mathbf{N}_{+}\right) .
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $T_{n}=\sum_{i=1}^{2 n}(-1)^{i+1} a_{i} a_{i+1}$, find $T_{n}$.
$$
\begin{array}{l}
a_{n+1}=f\left(\frac{1}{a_{n}}\right)=\frac{2+3 a_{n}}{3}=a_{n}+\frac{2}{3} \\
\Rightarrow a_{n}=\frac{2}{3} n+\frac{1}{3} . \\
\text { (2) } T_{n}=\sum_{i=1}^{2 n}(-1)^{i+1} a_{i} a_{i+1} \\
=a_{2}\left(a_{1}-a_{3}\right)+a_{4}\left(a_{3}-a_{5}\right)+\cdots+ \\
\quad a_{2 n}\left(a_{2 n-1}-a_{2 n+1}\right) \\
=-\frac{4}{3}\left(a_{2}+a_{4}+\cdots+a_{2 n}\right) \\
=-\frac{4}{3} \cdot \frac{n\left(\frac{5}{3}+\frac{4 n}{3}+\frac{1}{3}\right)}{2} \\
=-\frac{4}{9}\left(2 n^{2}+3 n\right) .
\end{array}
$$