Four. (50 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \\
a_{n}=3 a_{n-1}-a_{n-2}\left(n \in \mathbf{Z}_{+}, n \geqslant 3\right) .
\end{array}
$$

Does there exist a positive integer $n$ such that $2^{2016} \| a_{n}$, $\left(2^{2016} \mid a_{n}\right.$, and $\left.2^{2017} \nmid a_{n}\right)$? If it exists, find the smallest positive integer $n$; if not, explain the reason.
There exist infinitely many positive integers $n$ such that $2^{2016} \| a_{n}$, and the smallest positive integer $n$ is $3 \times 2^{2013}$.
If $2^{a} \| m$, then we denote $v_{2}(m)=a$.
Therefore, $v_{2}(m n)=v_{2}(m)+v_{2}(n)$, and
$v_{2}(m+n) \geqslant \min \left\{v_{2}(m), v_{2}(n)\right\}$.
If $v_{2}(m) \neq v_{2}(n)$, then
$v_{2}(m+n)=\min \left\{v_{2}(m), v_{2}(n)\right\}$.
From $a_{n}=3 a_{n-1}-a_{n-2} \equiv a_{n-1}-a_{n-2}(\bmod 2)$,

we know that $\left\{a_{n}\right\}$ is a periodic sequence with period 3 modulo 2.
Therefore, when $3 \uparrow n$, $v_{2}\left(a_{n}\right)=0$.
Let $n=3 k(k \geqslant 1)$. Clearly,
$$
\begin{array}{l}
a_{n}=\frac{1}{\sqrt{5}}\left(\left(\frac{3+\sqrt{5}}{2}\right)^{n}-\left(\frac{3-\sqrt{5}}{2}\right)^{n}\right) \\
\Rightarrow a_{3 k}=\frac{1}{\sqrt{5}}\left(\left(\frac{3+\sqrt{5}}{2}\right)^{3 k}-\left(\frac{3-\sqrt{5}}{2}\right)^{3 k}\right) \\
=\frac{1}{\sqrt{5}}\left((9+4 \sqrt{5})^{k}-(9-4 \sqrt{5})^{k}\right) \\
=2 \sum_{0 \leqslant i \leqslant\left[\frac{k-1}{2}\right]} \mathrm{C}_{k}^{2 i+1} 9^{k-(2 i+1)} \times 4^{2 i+1} \times 5^{i} \text {. } \\
\text { Also, }(2 i+1) \mathrm{C}_{k}^{2 i+1}=k \mathrm{C}_{k-1}^{2 i} \\
\Rightarrow v_{2}\left(\mathrm{C}_{k}^{2 i+1}\right) \geqslant v_{2}(k)(i \geqslant 1) \\
\Rightarrow v_{2}\left(a_{3 k}\right)=3+v_{2}(k) \text {. } \\
\end{array}
$$

Thus, $v_{2}\left(a_{3 k}\right)=2016$ if and only if $3+v_{2}(k)=2016$, i.e., $v_{2}(k)=2013$.

Therefore, $2^{2016} \| a_{n}$ if and only if $n=3 \times 2^{2013} m$, where $(m, 2)=1$.
In summary, the smallest positive integer $n$ is $3 \times 2^{2013}$.