During the Teachers' Day, 200 teachers at a school sent text messages to greet each other, with each teacher sending exactly 1 text message to another teacher. Now, from them, the maximum number of $k$ teachers can be selected to attend an award ceremony, such that none of them has sent a text message to any of the others. Try to find the minimum value of $k$.
Let the 200 teachers be denoted as $A_{1}, A_{2}, \cdots, A_{200}$, and let $G=\left\{A_{1}, A_{2}, \cdots, A_{200}\right\}$. If $A_{i}$ sends a message to $A_{j}$, then we mark an arrow from $A_{i}$ to $A_{j}$ as
$A_{i} \rightarrow A_{j}(1 \leqslant i \neq j \leqslant 200)$.
First, construct an instance for $k=67$.
In $G$, let
$$
A_{1} \rightarrow A_{2} \rightarrow A_{1}, A_{3 i} \rightarrow A_{3 i+1} \rightarrow A_{3 i+2} \rightarrow A_{3 i} \text {, }
$$

resulting in 1 pair of two people and 66 triangles $\triangle A_{3 i} A_{3 i+1} A_{3 i+2}$. From the pair, we can select one person, and from each triangle, we can select one person $A_{3 i}(i=1,2, \cdots, 66)$, hence
$$
A=\left\{A_{1}, A_{3}, A_{6}, \cdots, A_{195}, A_{198}\right\}
$$

satisfies the requirement.
Next, prove that for the largest $k$ that satisfies the requirement, $k \geqslant 67$.

Assume $A=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$ is a set of $k$ representatives that satisfy the requirement, and let
$$
B=G \backslash A=\left\{A_{k+1}, A_{k+2}, \cdots, A_{200}\right\} .
$$

Next, calculate the number of arrows in $G$.
On one hand, by the problem statement, $G$ has exactly 200 arrows.
On the other hand, estimate the number of arrows in $G$.
Note that $G=A \cup B$, and consider two cases.
(1) The number of arrows within $A$ is 0.
(2) There are at least $200-k$ arrows between $A$ and $B$ (since each person in $B$ must have at least one arrow to someone in $A$). Otherwise, if there exists a person $A_{j}(k+1 \leqslant j \leqslant 200)$ in $B$ who has no arrow to anyone in $A$, then $A^{\prime}=A \cup\left\{A_{j}\right\}$ also satisfies the requirement, but $\left|A^{\prime}\right|>k$, contradicting the maximality of $k$.
(3) There are at least $200-2 k$ arrows among the people in $B$.

In fact, partition $B$ into $m$ non-empty subsets $B_{1}$, $B_{2}, \cdots, B_{m}$, such that any two people in each $B_{i}$ can be connected by a sequence of arrows. By the maximality of $k$, we have $m \leqslant k$. By the definition of $B_{i}$, each $B_{i}$ has at least $\left|B_{i}\right|-1$ $(i=1,2, \cdots, m)$ arrows. Therefore, $B$ has at least
$$
\sum_{i=1}^{m}\left(\left|B_{i}\right|-1\right)=(200-k)-m \geqslant 200-2 k
$$

arrows.
Thus, $G$ has at least
$$
(200-k)+(200-2 k)=400-3 k
$$

arrows. Therefore,
$$
400-3 k \leqslant 200 \Rightarrow 3 k \geqslant 200 \Rightarrow k \geqslant 67 \text {. }
$$

In summary, the minimum value of $k$ is 67.