For nine non-negative real numbers $a_{1}, a_{2}, \cdots, a_{9}$ whose sum is 1, let
$$
\begin{array}{l}
S=\min \left\{a_{1}, a_{2}\right\}+2 \min \left\{a_{2}, a_{3}\right\}+\cdots+8 \min \left\{a_{8}, a_{9}\right\}+9 \min \left\{a_{9}, a_{1}\right\}, \\
T=\max \left\{a_{1}, a_{2}\right\}+2 \max \left\{a_{2}, a_{3}\right\}+\cdots+8 \max \left\{a_{8}, a_{9}\right\}+9 \max \left\{a_{9}, a_{1}\right\},
\end{array}
$$

where $\min \{x, y\}$ denotes the smaller of $x, y$, and $\max \{x, y\}$ denotes the larger of $x, y$. Let the maximum possible value of $S$ be $S_{0}$. When $S=S_{0}$, find all possible values of $T$.
Apply the weighted mean inequality: $x, y, \geqslant 0, m, n>0 \Rightarrow \min \{x, y\} \leqslant \frac{m x+n y}{m+n}$.
$$
\begin{array}{l}
\text { Then } S=\min \left\{a_{1}, a_{2}\right\}+2 \min \left\{a_{2}, a_{3}\right\}+\cdots+8 \min \left\{a_{8}, a_{9}\right\}+9 \min \left\{a_{9}, a_{1}\right\} \\
\leqslant a_{2}+2 a_{2}+3 a_{3}+4 a_{4}+5 a_{5}+6 a_{6}+\left(6 a_{7}+a_{8}\right)+\left(5 a_{8}+3 a_{9}\right)+\left(3 a_{9}+6 a_{1}\right) \\
\leqslant 6\left(a_{1}+a_{2}+\cdots+a_{9}\right)=6,
\end{array}
$$

Equality holds when $a_{2}=a_{3}=a_{4}=a_{5}=0, a_{6} \leqslant a_{7}$ and $a_{7}=a_{8}=a_{9}=a_{1}$. Let $a_{6}=x \Rightarrow a_{7}=a_{8}=a_{9}=a_{1}=\frac{1-x}{4}$, then $x \leqslant \frac{1-x}{4} \Rightarrow x \in\left[0, \frac{1}{5}\right]$. Then $T=\max \left\{a_{1}, a_{2}\right\}+2 \max \left\{a_{2}, a_{3}\right\}+\cdots+8 \max \left\{a_{8}, a_{9}\right\}+9 \max \left\{a_{9}, a_{1}\right\}$ $=a_{1}+5 a_{6}+(6+7+8+9) a_{1}=5 x+\frac{31-31 x}{4}=\frac{31-11 x}{4} \in\left[\frac{36}{5}, \frac{31}{4}\right]$. Therefore, the range of $T$ is $\left[\frac{36}{5}, \frac{31}{4}\right]$.