4. A manager of a company has eight employees under him. One day, the manager convenes an employee meeting, with the following rules:
(1) Each employee meeting lasts 1 hour, and there are no breaks between meetings;
(2) Each employee meeting is attended by three employees called by the manager;
(3) Any two employees must have attended at least one meeting together;
(4) Each employee cannot leave the company from the time they attend their first meeting to the time they attend their last meeting.

For all meeting arrangements that meet the requirements, consider the employee who stays in the company the longest on that day. What is the minimum number of hours this employee stays in the company? Prove your conclusion.
4. The shortest is 9 hours.

First, provide a construction to show that 9 hours can be achieved. Let $1 \sim 8$ represent eight employees, and the order of meetings is
$$
(1,2,3),(1,2,4),(1,5,7),(1,6,8) \text {, }
$$
$(3,4,5),(3,4,6),(2,5,8),(2,6,7),(3,7,8),(4,7,8),(5,6,8)$.
Next, prove that there must be an employee who worked 9 hours.
Consider two employees participating in the same meeting as one interaction.

Since each employee can only interact with two other employees in one meeting, each employee must have attended at least 4 meetings. Therefore, the total number of meetings is at least $\left\lceil\frac{4 \times 8}{3}\right\rceil=11$ times, where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.

Now consider the first two meetings. Since two meetings with the same participants are meaningless, we can remove these meetings. Thus, at least four members participated in the first two meetings, denoted as set $A$. At least four members also participated in the last two meetings, denoted as set $B$.

If $A$ and $B$ have common members, then he must have worked at least 9 hours.
If $A$ and $B$ have no common members, then it must be that
$$
|A|=|B|=4 \text {. }
$$

Consider the first meeting with members from $B$ to the last meeting with members from $A$. Thus, the other meetings are all at the beginning or end of the day, and are composed solely of members from $A$ or $B$. Since $A$ and $B$ each have four members, they need to interact 16 times. For each meeting, there must be two people from the same group of $A$ or $B$, so they can only interact twice at most. Therefore, there must be at least eight such meetings. Hence, in these meetings, the member from $B$ who participated in the first one must have worked at least 9 hours.

Therefore, there must be a member who worked 9 hours on that day.