9. (14 points) Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=1+2+\cdots+n\left(n \in \mathbf{N}_{+}\right) \text {, }
$$

take all multiples of 3 from this sequence to form a new sequence $b_{1}, b_{2}, \cdots, b_{m}, \cdots$. Find the sum of the first $2 m$ terms of the sequence $\left\{b_{m}\right\}$, denoted as $S_{2 m}$.
Obviously, $a_{n}=\frac{n(n+1)}{2}$.
If $a_{n}=3 t\left(t \in \mathbf{Z}_{+}\right)$, then since $n$ and $n+1$ are coprime, one of $n$ and $n+1$ must be a multiple of 3.
When $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $a_{3 k}=\frac{3 k(3 k+1)}{2}$;
When $n+1=3 k$, i.e., $n=3 k-1\left(k \in \mathbf{Z}_{+}\right)$,
$$
a_{3 k-1}=\frac{3 k(3 k-1)}{2} \text {. }
$$

Then, in the sequence $\left\{a_{n}\right\}$, among the three consecutive terms $a_{3 k-2}, a_{3 k-1}, a_{3 k}$, there are two consecutive terms $a_{3 k-1}, a_{3 k}$ that are multiples of 3.
Thus, $b_{1}=a_{2}, b_{2}=a_{3}, b_{3}=a_{5}, b_{4}=a_{6}$,
$$
\begin{array}{l}
b_{2 m-1}=a_{3 m-1}, b_{2 m}=a_{3 m} . \\
S_{2 m}=\sum_{k=1}^{m}\left(a_{3 k-1}+a_{3 k}\right)=\sum_{k=1}^{m} 9 k^{2} \\
=\frac{3}{2} m(m+1)(2 m+1) .
\end{array}
$$