$5 \cdot 113$ Suppose there are 30 people sitting around a round table, each of whom is either a fool or a wise person. Each person is asked: “Is your right-hand neighbor a wise person or a fool?” Wise people always give the correct answer, while fools may answer either correctly or incorrectly. It is known that the number of fools does not exceed $F$. Find the largest $F$ such that it is always possible to identify at least one wise person.
[Solution] Let the number of fools be $F \neq 0$. We divide all the people sitting around the round table into groups as follows: smart people sitting next to each other form one group; fools sitting next to each other form another group. Thus, we get $2k$ groups, with $k$ groups of smart people and $k$ groups of fools, alternating around the table. Let the number of smart people in the $i$-th group be $w_i$, and the number of fools in the $i$-th group be $f_i$, for $i=1,2, \cdots, k$. Clearly, we have
$$
\begin{array}{c}
w_{1}+w_{2}+\cdots+w_{k}+f_{1}+f_{2}+\cdots+f_{k}=30, \\
f_{1}+f_{2}+\cdots+f_{k}=F .
\end{array}
$$

Consider the sequence of people who are called smart by their left neighbor. Clearly, the number of terms in the sequence formed by a group of smart people ($w_i$) is no less than $w_i - 1$, and as long as one term in the sequence is a smart person, all the people following him are also smart. The length of the sequence formed by a group of fools does not exceed $f_i - 1$. Therefore, when $w_{i_0} = \frac{\max w_i}{i} > \max f_i$, the number of terms in the sequence formed by these $w_{i_0}$ smart people is no less than any $f_i$, so this sequence cannot consist solely of fools, and the last person in the sequence must be a genuine smart person. At this point, we have
$$
\begin{array}{l}
\max w_i \geq \frac{30-F}{k}, \\
\max f_i \leq F - k + 1, \\
\frac{30-F}{k} > F - k + 1, \quad k=1,2, \cdots, F, \\
\end{array}
$$

That is, inequality (2) holds for all $k$ when we can identify one smart person. Inequality (2) is equivalent to
$$
k^2 - (F+1)k + 30 - F > 0
$$

for all $1 \leq k \leq F$, and this is only necessary if
$$
\Delta = (F+1)^2 + 4(F-30) < 0.
$$

Solving (3) gives
$$
F < -3 + \sqrt{128} < -3 + 12 = 9.
$$

Thus, when $F \leq 8$, we can identify one smart person based on the given answers.
On the other hand, when $F = 9$, the above conclusion no longer holds, meaning it is not always possible to identify one smart person. For this, we consider the 30 people shown in the figure below, where black dots represent fools and small circles represent smart people, with a total of 9 fools. In the figure, people with $f$ are called fools by their left neighbor, while others are called smart. It is easy to see that when the figure is rotated $60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ$ counterclockwise around the center, the answers remain unchanged, but in reality, each position can be either a smart person or a fool. Therefore, it is impossible to identify any one genuine smart person based on the answers.
In summary, the maximum value of $F$ is 8.