11. (20 points) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
a_{1}=3, b_{1}=1
$$

and for any $n \in \mathbf{Z}_{+}$, we have
$$
\left\{\begin{array}{l}
a_{n+1}=a_{n}+b_{n}+\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}} \\
b_{n+1}=a_{n}+b_{n}-\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}}
\end{array} .\right.
$$
(1) Find the general terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$;
(2) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $S_{n}=\sum_{i=1}^{n}\left[a_{i}\right], T_{n}=\sum_{i=1}^{n}\left[b_{i}\right]$. Find the smallest $n \in \mathbf{Z}_{+}$ such that
$$
\sum_{k=1}^{n}\left(S_{k}+T_{k}\right)>2017 .
$$
11. (1) From the problem, we have
$$
\begin{array}{l}
a_{n+1}+b_{n+1}=2\left(a_{n}+b_{n}\right), \\
a_{n+1} b_{n+1}=3 a_{n} b_{n} . \\
\text { Also, } a_{1}+b_{1}=4, a_{1} b_{1}=3 \text {, then } \\
a_{n}+b_{n}=\left(a_{1}+b_{1}\right) 2^{n-1}=2^{n+1}, \\
a_{n} b_{n}=a_{1} b_{1} 3^{n-1}=3^{n} .
\end{array}
$$

Notice that, $a_{n} 、 b_{n}>0$ and $a_{n}>b_{n}$.
Thus, $a_{n}=2^{n}+\sqrt{4^{n}-3^{n}}, b_{n}=2^{n}-\sqrt{4^{n}-3^{n}}$.
(2) From (1), we know $a_{n}+b_{n}=2^{n+1}$.
Then $2^{n+1}-2<\left[a_{n}\right]+\left[b_{n}\right] \leqslant 2^{n+1}$.
We will prove by contradiction that when $n \geqslant 2$, there does not exist $n$ such that $\sqrt{4^{n}-3^{n}} \in \mathbf{Z}_{+}$.
If there exist $n 、 t \in \mathbf{Z}_{+}$ such that $4^{n}-3^{n}=t^{2}$, then
$$
\left(2^{n}+t\right)\left(2^{n}-t\right)=3^{n} \text {. }
$$

When 3 divides $\left(2^{n}+t\right)$ and 3 divides $\left(2^{n}-t\right)$, $3 \mid t$, which implies $3 \mid 4^{n}$, a contradiction.
Thus, only $2^{n}-t=1 \Rightarrow t=2^{n}-1$.
Substituting into equation (1) gives $2^{n+1}=3^{n}+1$.
However, $2^{n+1}-3^{n}=8 \times 2^{n-2}-9 \times 3^{n-2}<0$, a contradiction.
Therefore, only when $n=1$, $a_{n} 、 b_{n} \in \mathbf{Z}_{+}$, at this time, $a_{1}+b_{1}=4$;
When $n \geqslant 2$, $\left[a_{n}\right]+\left[b_{n}\right]=2^{n+1}-1$.
Then $S_{n}+T_{n}=4+\sum_{i=2}^{n}\left(2^{i+1}-1\right)$ $=2^{n+2}-n-3$.
$$
\text { Hence } \sum_{k=1}^{n}\left(S_{k}+T_{k}\right)=2^{n+3}-8-\frac{n(n+1)}{2}-3 n \text {. }
$$

In particular, when $n=8$,
$$
\sum_{k=1}^{8}\left(S_{k}+T_{k}\right)=2^{11}-8-\frac{8(8+1)}{2}-24=1980 \text {. }
$$

Therefore, the smallest $n=9$.