Four, (50 points) During the Teachers' Day at a certain school, 200 teachers sent text messages to greet each other, with each person sending exactly one message to another teacher. Now, from them, the maximum number of $k$ representatives can be selected to attend a commendation conference, such that none of them has sent a message to any of the others. Try to find the minimum value of $k$.

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The minimum value of $k$ sought is 67.
Let the 200 teachers be denoted as $A_{1}, A_{2}, \cdots, A_{200}$.
If $A_{i}$ sends a message to $A_{j}$, then mark $A_{i}$ and $A_{j}$ with an arrow as $A_{i} \rightarrow A_{j}(1 \leqslant i \neq j \leqslant 200)$. Let $G=\left\{A_{1}, A_{2}, \cdots, A_{200}\right\}$.
First, construct an instance for $k=67$.
In $G$, let
$$
A_{1} \rightarrow A_{2} \rightarrow A_{1}, A_{3 i} \rightarrow A_{3 i+1} \rightarrow A_{3 i+2} \rightarrow A_{3 i},
$$

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, one person can be selected, and from each triangle, exactly one person $A_{3 i}(i=1,2, \cdots, 66)$ can be selected, hence
$$
A=\left\{A_{1}, A_{3}, A_{6}, \cdots, A_{195}, A_{198}\right\}
$$

satisfies the requirement, and the maximum $k=67$.
Next, we prove that for any $k$ that satisfies the requirement, $k \geqslant 67$.
We will prove that $k \leqslant 66$ does not hold, using proof by contradiction.
Assume the maximum number of selected people is $k$, denoted as $A_{1}, A_{2}, \cdots, A_{k}$, and the remaining $200-k$ people are denoted as $B_{1}, B_{2}, \cdots, B_{200-k}$.

It is easy to see that for any $B_{i}$, there exists an $A_{j}$ such that $A_{j}$ has sent a message to $B_{i}$ or $B_{i}$ has sent a message to $A_{j}$; otherwise, let $B_{i}$ be $A_{k+1}$, then $A_{1}, A_{2}, \cdots, A_{k}, A_{k+1}$ would satisfy the condition, contradicting the maximality of $k$. And $A_{1}, A_{2}, \cdots, A_{k}$ can send messages to at most $k$ of the $B_{1}, B_{2}, \cdots, B_{200-k}$ people.
Thus, among $B_{1}, B_{2}, \cdots, B_{200-k}$, there are at least
$$
200-k-k=200-2 k
$$

people who have sent messages to someone in $A_{1}, A_{2}, \cdots, A_{k}$. These $200-2 k$ people have not sent messages to each other, and $k \leqslant 66$.

Therefore, $200-2 k \geqslant 68 > k$, which contradicts the maximality of $k$.
In conclusion, the minimum value of $k$ is 67.