Four. (50 points) A country has built a time machine, which resembles a circular subway track, with 2014 stations (numbered $1, 2, \cdots, 2014$) evenly distributed along it, each corresponding to a year. The starting and ending stations are both the 1st station (corresponding to the year 2014). To save costs, the machine stops at only half of the stations each time it runs a full circle. Due to technical reasons, it must stop at least once every three stations, and no two stations it stops at can be diametrically opposite on the circular track. Determine the number of different ways to stop at the stations.
Let $2 n=2014$. Suppose there are $f(n)$ different ways of stopping.

First, for each way of stopping, define the stations where the machine stops as class $a$ stations, and the stations where it does not stop as class $b$ stations. Then, class $a$ stations and class $b$ stations are paired one-to-one, forming diametrically opposite points.

Obviously, there cannot be three consecutive class $a$ stations (otherwise, if $A$, $B$, and $C$ are three consecutive class $a$ stations, then their diametrically opposite points $A'$, $B'$, and $C'$ would be three consecutive class $b$ stations, and the machine would not stop at these stations, contradicting the problem statement), and the 1st station is a class $a$ station, and the $(n+1)$-th station is a class $b$ station.

Thus, each way of stopping corresponds to a classification of the 1st to $n$-th stations (the classification of the $(n+1)$-th to $2n$-th stations is uniquely determined by the classification of the 1st to $n$-th stations), such that no three consecutive stations are of the same type.

Next, consider the classification of $k+1\left(k \in \mathbf{Z}_{+}\right)$ consecutive stations (where the first and last stations are class $a$, and no three consecutive stations are of the same type). Let the number of such classification ways be $g(k)$.
$$
\begin{array}{l}
\text { Clearly, } g(1)=1, g(2)=1, g(3)=2^{2}-1=3, \\
g(4)=2^{3}-4=4.
\end{array}
$$

For $k \geqslant 5$, consider the number of class $b$ stations between the last two class $a$ stations, denoted as $t$. If $t=2$, the number of classification ways is $g(k-3)$; if $t=1$, the number of classification ways is $g(k-2)$; if $t=0$, the adjacent station is class $b$, and the number of classification ways is $g(k-3)+g(k-4)$.
$$
\begin{array}{l}
\text { Hence, } g(k)=g(k-2)+2 g(k-3)+g(k-4). \\
\text { Let } h(k)=g(k)-g(k-3). \\
\text { Then, } h(k)=g(k-2)+g(k-3)+g(k-4) \\
=h(k-2)+g(k-5)+g(k-3)+g(k-4) \\
=h(k-2)+h(k-1).
\end{array}
$$

From the conclusion (1) and equation (2), we get $g(0)=1$.
$$
\begin{array}{l}
\text { Then, } h(3)=g(3)-g(0)=2, \\
h(4)=g(4)-g(1)=3.
\end{array}
$$

From equation (4), $h(k)$ is the $k$-th Fibonacci number $f_{k}$, i.e.,
$$
h(k)=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k}-\left(\frac{1-\sqrt{5}}{2}\right)^{k}\right].
$$

From equation (3), we get
$$
\begin{array}{l}
g(3 k)=g(0)+\sum_{i=1}^{k}(g(3 i)-g(3 i-3)) \\
=1+\sum_{i=1}^{k} f_{3 i}, \\
g(3 k+1)=g(1)+\sum_{i=1}^{k}(g(3 i+1)-g(3 i-2)) \\
=1+\sum_{i=1}^{k} f_{3 i+1}, \\
g(3 k+2)=g(2)+\sum_{i=1}^{k}(g(3 i+2)-g(3 i-1)) \\
=1+\sum_{i=1}^{k} f_{3 i+2}.
\end{array}
$$

Next, calculate the number of classification ways for the 1st to $n$-th stations, $f(n)$.
If $n$ is class $a$, the corresponding number of classification ways is $g(n-1)$.
If $n$ is class $b$, then $n-1$, $n+2$ are class $a$, and 2 is class $b$.
If 3 is class $a$, the corresponding number of classification ways is $g(n-4)$;
If 3 is class $b$, then 4 is class $a$, and the corresponding number of classification ways is $g(n-5)$.
Thus, $f(n)=g(n-1)+g(n-4)+g(n-5)$.
Finally, find $f(1007)$.
$$
\begin{array}{l}
f(1007)=g(1006)+g(1003)+g(1002) \\
=3+\sum_{i=1}^{335} f_{3 i+1}+\sum_{i=1}^{334} f_{3 i+1}+\sum_{i=1}^{344} f_{3 i} \\
=3+\sum_{i=1}^{335} f_{3 i+1}+\sum_{i=1}^{334} f_{3 i+2}=3+f_{1006}+\sum_{i=1}^{344} f_{3 i+3} \\
=1+f_{1006}+\sum_{i=1}^{335} f_{3 i}.
\end{array}
$$
By
$$
\left\{\begin{array}{l}
\sum_{i=1}^{k} f_{3 i}+\sum_{i=1}^{k} f_{3 i+1}=\sum_{i=1}^{k} f_{3 i+2} \\
\sum_{i=1}^{k} f_{3 i}+\sum_{i=1}^{k} f_{3 i-1}=\sum_{i=1}^{k} f_{3 i+1}
\end{array}\right.
\Rightarrow \sum_{i=1}^{k} f_{3 i}=\frac{1}{2}\left(f_{3 k+2}-f_{2}\right).
$$

Thus, $f(1007)=1+f_{1006}+\frac{1}{2}\left(f_{1007}-f_{2}\right)$
$$
\begin{aligned}
= & \frac{1}{2}\left(1+2 f_{1006}+f_{1007}\right) \\
= & \frac{1}{2}\left\{1+\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{1006}\left(2+\frac{1+\sqrt{5}}{2}\right)-\right.\right. \\
& \left.\left.\left(\frac{1-\sqrt{5}}{2}\right)^{1006}\left(2+\frac{1-\sqrt{5}}{2}\right)\right]\right\} \\
= & \frac{1}{2}\left[1+\left(\frac{1+\sqrt{5}}{2}\right)^{1007}+\left(\frac{1-\sqrt{5}}{2}\right)^{1007}\right].
\end{aligned}
$$
(Song Qiang, 27 Jiaqingli, Hongqiao District, Tianjin, 300134)