4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=1, a_{2}=3, a_{n}=3 a_{n-1}-a_{n-2}, n \in \mathbf{N}^{*}, n \geqslant 3$. Does there exist a positive integer $n$ such that $2^{2016} \| a_{n}$ (i.e., $2^{2016}$ divides $a_{n}$, but $2^{2017}$ does not divide $a_{n}$)? If it exists, find the smallest positive integer $n$; if not, explain why.
If $2^{a} \| m$, then we denote $v_{2}(m)=a$. Clearly, $v_{2}(m n)=v_{2}(m)+v_{2}(n)$, and $v_{2}(m+n) \geqslant \min \left\{v_{2}(m), v_{2}(n)\right\}$; if $v_{2}(m) \neq v_{2}(n)$, then $v_{2}(m+n)=\min \left\{v_{2}(m), v_{2}(n)\right\}$.
Claim: (1) $v_{2}\left(a_{n}\right)=\left\{\begin{array}{l}0, n \not \equiv 0(\bmod 3), \\ 3+v_{2}(k), n=3 k .\end{array}\right.$ The table is as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & $\cdots$ \\
\hline$a_{n}(\bmod 2)$ & 1 & 1 & 0 & 1 & 1 & 0 & 1 & 1 & 0 & $\cdots$ \\
\hline
\end{tabular}

Therefore, when $n \not \equiv 0(\bmod 3)$, we have $v_{2}\left(a_{n}\right)=0$.
The characteristic equation of the sequence $\left\{a_{n}\right\}$ is $x^{2}-3 x+1=0 \Rightarrow x_{1,2}=\frac{3 \pm \sqrt{5}}{2}$,
Thus, $a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{3+\sqrt{5}}{2}\right)^{n}-\left(\frac{3-\sqrt{5}}{2}\right)^{n}\right]$
$\Rightarrow a_{3 k}=\frac{1}{\sqrt{5}}\left[\left(\frac{3+\sqrt{5}}{2}\right)^{3 k}-\left(\frac{3-\sqrt{5}}{2}\right)^{3 k}\right]=\frac{1}{\sqrt{5}}\left[(9+4 \sqrt{5})^{k}-(9-4 \sqrt{5})^{k}\right]$
$=\frac{1}{\sqrt{5}}\left[\sum_{i=0}^{k} C_{k}^{i} 9^{k-i}(4 \sqrt{5})^{i}-\sum_{i=0}^{k} C_{k}^{i} 9^{k-i}(-4 \sqrt{5})^{i}\right]=\frac{2}{\sqrt{5}} \sum_{i=0}^{\left[\frac{k-1}{2}\right]} C_{k}^{2 i+1} 9^{k-2 i-1}(4 \sqrt{5})^{2 i+1}$
$=2 \sum_{i=0}^{\left[\frac{k-1}{2}\right]} C_{k}^{2 i+1} \cdot 9^{k-2 i-1} \cdot 4^{2 i+1} \cdot 5^{i}=\sum_{i=0}^{\left[\frac{k-1}{2}\right]} C_{k}^{2 i+1} \cdot 2^{4 i+3} \cdot 3^{2 k-4 i-2} \cdot 5^{i}$
$=C_{k}^{1} \cdot 2^{3} \cdot 3^{2 k-2}+C_{k}^{3} \cdot 2^{7} \cdot 3^{2 k-6} \cdot 5+\cdots+C_{k}^{2 i+1} \cdot 2^{4 i+3} \cdot 3^{2 k-4 i-2} \cdot 5^{i}+\cdots$.
Since $C_{k}^{2 i+1}=\frac{k!}{(2 i+1)!(k-2 i-1)!}=\frac{k}{2 i+1} \cdot \frac{(k-1)!}{(2 i)!(k-2 i-1)!}=\frac{k}{2 i+1} C_{k-1}^{2 i}$
$$
\Rightarrow(2 i+1) C_{k}^{2 i+1}=k C_{k-1}^{2 i} \Rightarrow v_{2}\left(C_{k}^{2 i+1}\right)=v_{2}(k)+v_{2}\left(C_{k-1}^{2 i}\right), i \geqslant 0 .
$$

In summary, we have $v_{2}\left(a_{3 k}\right)=3+v_{2}(k)$.
(2) $v_{2}\left(a_{3 k}\right)=2016$ if and only if $3+v_{2}(k)=2016 \Rightarrow v_{2}(k)=2013$,

Thus, $2^{2016} \| a_{n} \Leftrightarrow n=3 \cdot 2^{2013} \cdot m$, where $(m, 2)=1$.
In summary, the smallest positive integer $n=3 \cdot 2^{2013}$.