9.4. A group of scientists work at a research institute. During an 8-hour workday, each scientist visits the cafeteria at least once. It is known that for any two scientists, the total time that at least one of them is in the cafeteria is at least $x(x>4)$ hours. Find the maximum possible number of scientists working at the institute (in terms of $x$).
9.4.2 $\left[\frac{x}{2 x-8}\right]$.

Suppose there are $n$ scientists in the institute, $t_{ij}$ represents the time when exactly one of the $i$-th and $j$-th scientists is in the coffee shop, and let $S=\sum_{1 \leqslant i<j \leqslant n} t_{ij}$. Then $S \geqslant \mathrm{C}_{n}^{2} x$.

On the other hand, divide the 8-hour working time into finite segments $t_{1}, t_{2}, \cdots, t_{m}$, such that no scientist enters or leaves the coffee shop during each segment. Let $k_{i}$ be the number of scientists in the coffee shop during the segment $t_{i}$. Then
$$
\begin{array}{l}
S=\sum_{i=1}^{m} t_{i} k_{i}\left(n-k_{i}\right) \leqslant \sum_{i=1}^{m} t_{i}\left[\frac{n}{2}\right]\left(n-\left[\frac{n}{2}\right]\right) \\
=8\left[\frac{n}{2}\right]\left(n-\left[\frac{n}{2}\right]\right) . \\
\text { Thus, } 8\left[\frac{n}{2}\right]\left(n-\left[\frac{n}{2}\right]\right) \geqslant \frac{n(n-1)}{2} x .
\end{array}
$$

If $n$ is even, let $n=2 l$, then
$$
8 l^{2} \geqslant l(2 l-1) x \text {, }
$$

which implies $l \leqslant \frac{x}{2 x-8}, n \leqslant 2\left[\frac{x}{2 x-8}\right]$;
If $n$ is odd, let $n=2 l+1$, then
$$
8 l(l+1) \geqslant l(2 l+1) x,
$$

which implies $l \leqslant \frac{8-x}{2 x-8}$,
$$
n \leqslant 2\left[\frac{8-x}{2 x-8}\right]+1=2\left[\frac{x}{2 x-8}\right]-1 .
$$

Therefore, in all cases, $n \leqslant 2\left[\frac{x}{2 x-8}\right]$.
The following example shows that the inequality can be achieved.
Let $n=2 l=2\left[\frac{x}{2 x-8}\right]$. Let $K=C_{n}^{l}$.
Divide the 8-hour working time into $K$ equal segments, each segment corresponding to $l$ out of the $n$ scientists, and different segments correspond to different sets of $l$ scientists. By symmetry, for any two scientists, the total time when exactly one of them is in the coffee shop is the same, denoted as $y$. Then
$$
8 l(2 l-l)=S=\frac{2 l(2 l-1)}{2} y .
$$

Thus, $y=\frac{8 l}{2 l-1} \geqslant x$ (this inequality is equivalent to $l \leqslant \frac{x}{2 x-8}$).
Hence, the condition is satisfied.