1. A communication network consists of several terminals. If among any three terminals, at least two are directly connected, then this communication network is called "triple-connected." A communication network that satisfies the following conditions is referred to as a "windmill with $n$ blades."
$n$ pairs of terminals $\left\{x_{1}, y_{1}\right\},\left\{x_{2}, y_{2}\right\}, \cdots,\left\{x_{n}, y_{n}\right\}$ where $x_{i}$ is directly connected to $y_{i}(i=1,2, \cdots, n)$, and there exists a central terminal that is connected to all $2 n$ terminals $x_{1}, y_{1}, x_{2}, y_{2}, \cdots, x_{n}, y_{n}$.

Let the minimum number of terminals in any triple-connected communication network that contains a windmill with $n$ blades be denoted as $f(n)$. Find the value of $f(n)$.
1. Let each vertex in graph $G$ represent a terminal. If two terminals are directly connected, a red line is drawn between the corresponding vertices; otherwise, a blue line is drawn.
From the problem, we know that there are no blue triangles in graph $G$.
Let $|G|=m$.
(1) $n=1$.
A windmill with one blade is a red triangle.
(i) When $m \geqslant 6$, by Ramsey's theorem, we know that graph $G$ must contain a monochromatic triangle. Since there are no blue triangles in graph $G$, the monochromatic triangle must be red, i.e., $f(1) \leqslant 6$.
(ii) When $m=5$, consider a regular pentagon with its five sides colored red and its five diagonals colored blue. This graph is 3-connected and does not contain a red triangle. Therefore, $f(1) \geqslant 6$.
From (i) and (ii), we get $f^{\prime}(1)=6$.
(2) $n \geqslant 2$.
(i) If $m=4 n$, consider a graph composed of $4 n$ vertices $x_{1}, x_{2}, \cdots, x_{4 n}$, where the lines between $x_{1}, x_{2}, \cdots, x_{2 n}$ are all red, and the lines between $x_{2 n+1}, x_{2 n+2}, \cdots, x_{4 n}$ are all red, and the rest of the lines are blue. Then this graph is 3-connected and does not contain a windmill with $n$ blades.
Thus, $f(n) \geqslant 4 n+1$.
(ii) If $m=4 n-1$, we will prove that graph $G$ must contain a windmill with $n$ blades.

Take any point $x_{0}$ in graph $G$, let $B$ be the set of points connected to $x_{0}$ by red lines, and $C$ be the set of points connected to $x_{0}$ by blue lines. Clearly, any two points in set $C$ are connected by red lines; otherwise, if there are two points in set $C$ connected by a blue line, they would form a blue triangle with $x_{0}$, which is a contradiction.
If $|B| \geqslant 2 n+1$, let $B=\left\{x_{1}, x_{2}, \cdots, x_{2 n+1}\right\}$.
Then among $x_{1}, x_{2}, x_{3}$, there must be two points connected by a red line, say $x_{1} x_{2}$ is red. Hence, among $x_{3}, x_{4}, x_{5}$, there must be two points connected by a red line, say $x_{3} x_{4}$ is red, and so on, the edges $x_{1} x_{2}, x_{3} x_{4}, \cdots, x_{2 n-1} x_{n}$ are all red, forming a windmill with $n$ blades together with $x_{0}$.

If $|C| \geqslant 2 n+1$, since any two points in set $C$ are connected by red lines, set $C$ contains a red complete graph of order $2 n+1$, which clearly contains a windmill with $n$ blades.
$$
\begin{array}{l}
\text { If }|B|=|C|=2 n, \text { let } \\
B=\left\{x_{1}, x_{2}, \cdots, x_{2 n}\right\}, \\
C=\left\{y_{1}, y_{2}, \cdots, y_{2 n}\right\},
\end{array}
$$

and the $2 n$ points in set $C$ form a red complete graph of order $2 n$. If there exists a point $x_{1}$ in set $B$ connected to two points $y_{1}, y_{2}$ in set $C$ by red lines, then $y_{1}$ together with $\left\{x_{1}, y_{2}\right\},\left\{y_{3}, y_{4}\right\}$, $\cdots,\left\{y_{2 n-1}, y_{2 n}\right\}$ forms a windmill with $n$ blades. If any point in set $B$ is connected to at most one point in set $C$ by a red line, take any $x_{i}, x_{j} \in B(1 \leqslant i<j \leqslant 2 n)$, they are connected to at most two points in set $C$ by red lines. Take a point $y_{k}$ in set $C$ such that the edges $x_{i} y_{k}, x_{j} y_{k}$ are both blue, then the edge $x_{i} x_{j}$ is red, i.e., the $2 n$ points in set $B$ form a red complete graph of order $2 n$, and together with $x_{0}$ form a red complete graph of order $2 n+1$, which clearly contains a windmill with $n$ blades.
From (i) and (ii), we get $f(n)=4 n+1(n \geqslant 2)$.
In summary, $f(n)=\left\{\begin{array}{ll}6, & n=1 ; \\ 4 n+1, & n \geq 2 .\end{array}\right.$