$a_1,a_2,...,a_9$ are nonnegative reals with sum $1$. Define $S$ and $T$ as below:
$$S=\min\{a_1,a_2\}+2\min\{a_2,a_3\}+...+9\min\{a_9,a_1\}$$
$$T=\max\{a_1,a_2\}+2\max\{a_2,a_3\}+...+9\max\{a_9,a_1\}$$
When $S$ reaches its maximum, find all possible values of $T$.
1. Given the sequence \(a_1, a_2, \ldots, a_9\) of nonnegative reals with the sum \(a_1 + a_2 + \cdots + a_9 = 1\), we need to find the maximum value of \(S\) and the corresponding value of \(T\).

2. Define \(S\) and \(T\) as follows:
   \[
   S = \min\{a_1, a_2\} + 2\min\{a_2, a_3\} + \cdots + 9\min\{a_9, a_1\}
   \]
   \[
   T = \max\{a_1, a_2\} + 2\max\{a_2, a_3\} + \cdots + 9\max\{a_9, a_1\}
   \]

3. To maximize \(S\), we need to consider the terms \(\min\{a_i, a_{i+1}\}\). Notice that:
   \[
   S \leq a_2 + 2a_2 + 3a_3 + 4a_4 + 5a_5 + 6a_6 + 7a_7 + 8a_8 + 9a_9
   \]
   This inequality holds because \(\min\{a_i, a_{i+1}\} \leq a_{i+1}\).

4. Summing the terms, we get:
   \[
   S \leq 6a_1 + 3a_2 + 3a_3 + 4a_4 + 5a_5 + 6a_6 + 6a_7 + 6a_8 + 6a_9
   \]
   \[
   S \leq 6(a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9) = 6
   \]

5. The equality holds when \(a_2 = a_3 = a_4 = a_5 = 0\), \(a_7 = a_8 = a_9 = a_1 = t\), and \(a_6 \leq a_7\). Therefore, we have:
   \[
   a_1 + a_6 + a_7 + a_8 + a_9 = 1
   \]
   \[
   4t + a_6 = 1
   \]

6. Solving for \(a_6\):
   \[
   a_6 = 1 - 4t
   \]

7. To find \(T\), we substitute the values:
   \[
   T = t + 2a_6 + 3t + 4t + 5t + 6t + 7t + 8t + 9t
   \]
   \[
   T = t + 2(1 - 4t) + 3t + 4t + 5t + 6t + 7t + 8t + 9t
   \]
   \[
   T = t + 2 - 8t + 3t + 4t + 5t + 6t + 7t + 8t + 9t
   \]
   \[
   T = 11t + 2
   \]

8. The constraints on \(t\) are:
   \[
   1 - 4t \leq t \implies 1 \leq 5t \implies t \geq \frac{1}{5}
   \]
   \[
   1 - 4t \geq 0 \implies t \leq \frac{1}{4}
   \]

9. Therefore, \(t\) ranges from \(\frac{1}{5}\) to \(\frac{1}{4}\). Substituting these values into \(T\):
   \[
   T = 11 \left(\frac{1}{5}\right) + 2 = \frac{11}{5} + 2 = \frac{21}{5}
   \]
   \[
   T = 11 \left(\frac{1}{4}\right) + 2 = \frac{11}{4} + 2 = \frac{19}{4}
   \]

10. Thus, the possible values of \(T\) when \(S\) reaches its maximum are:
    \[
    T \in \left[\frac{21}{5}, \frac{19}{4}\right]
    \]

The final answer is \( \boxed{ \left[\frac{21}{5}, \frac{19}{4}\right] } \)