Given a positive integer $m$. Several honest people and $m$ ordinary people are sitting around a round table. Honest people never lie, while ordinary people will answer randomly. Person A comes to the scene to find an honest person as a friend, and everyone simultaneously tells A whether the next person in the clockwise direction is an honest person or an ordinary person. Question: What is the maximum number of honest people such that A cannot guarantee finding a friend based solely on their responses?
Let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$, and $x \rceil$ denote the smallest integer not less than the real number $x$.

When $k=\left\lfloor\frac{m}{2}\right\rfloor\left\lceil\frac{m}{2}\right\rceil+1$, it is possible that A cannot find a friend.

In the presence of $n(n>k)$ honest people and $m$ ordinary people, the following strategy can be adopted to find an honest person.

For any seating arrangement, a group of consecutive ordinary people (honest people) is defined as a group if and only if the people before and after them are all honest people (ordinary people). It is easy to see that the number of groups of honest people is the same as the number of groups of ordinary people.

Mark all the people around the table: if a person says the next person is honest, mark them as 1; otherwise, mark them as 0.

Note that the answers of a group of honest people will always be in the form $11 \cdots 10$.

Consider the longest sequence of $t$ consecutive people marked as $11 \cdots 10$ in the clockwise sequence around the table. If there are multiple such sequences, choose the one with the most consecutive 0s before it. Thus, the person marked as 0 in this sequence of $t$ people is an honest person.

Otherwise, these $t$ people are all ordinary people. We will discuss the following cases.
(1) There are two people marked as 0 before these $t$ people, i.e., there are $t+2$ people marked as $0011 \cdots 10$.
(i) If the person before these $t$ people is honest, then the person before him must be ordinary. Since there are $m$ ordinary people, the maximum number of consecutive groups of ordinary people is $m-t+1$, so the maximum number of consecutive groups of honest people is also $m-t+1$. Since the maximum number of consecutive honest people in each group is $t$, and there is at least one group with only one person, the number of honest people is
$$
n \leqslant(m-t) t+1 \leqslant\left\lfloor\frac{m}{2}\right\rfloor\left\lceil\frac{m}{2}\right\rceil+1<n,
$$
which is a contradiction.
(ii) If the person before these $t$ people is ordinary, then this group of ordinary people is at least $t+1$ long, so the maximum number of consecutive groups of ordinary people is $m-(t+1)+1$. Thus, the maximum number of consecutive groups of honest people is $m-t$. Since the maximum number of consecutive honest people in each group is $t$, the number of honest people is
$$
n \leqslant(m-t) t \leqslant\left\lfloor\frac{m}{2}\right\rfloor\left\lceil\frac{m}{2}\right\rceil<n,
$$
which is a contradiction.
(2) There is only one person marked as 0 before these $t$ people, then any sequence of $t$ consecutive people marked as $11 \cdots 10$ has only one person marked as 0 before it. Since there are $m$ ordinary people, the maximum number of consecutive groups of ordinary people is $m-t+1$.
(i) If there is no group of $t$ consecutive honest people, then the number of honest people is
$$
n \leqslant(m-t+1)(t-1) \leqslant\left\lfloor\frac{m}{2}\right\rfloor\left\lceil\frac{m}{2}\right\rceil<n.
$$
(ii) If there is a group of $t$ consecutive honest people, the person before them must be ordinary and marked as 0, and the person before him must be marked as 1, so the person before him must also be ordinary. Thus, the length of the group of ordinary people before each group of $t$ consecutive honest people is at least 2. Assuming there are $s$ groups of $t$ consecutive honest people, the number of groups of ordinary people is at most $m-t-2 s+s+1$. Therefore, $n \leqslant s t+(t-1)(m-t-2 s+s+1-s)$.
By $m<n$, we know that $t \geqslant 2, s \geqslant 1$. Thus,
$n \leqslant s t+(t-1)(m-t-2 s+s+1-s)$
$=(t-1)(m-t+1)+s(2-t)$
$\leqslant\left\lfloor\frac{m}{2}\right\rfloor\left\lceil\frac{m}{2}\right\rceil<n$,
which is a contradiction.
In summary, when there are $n$ honest people and $m$ ordinary people, the person found must be an honest person.

Next, we prove: When $k=\left\lfloor\frac{m}{2}\right\rfloor\left\lceil\frac{m}{2}\right\rceil+1$, it is possible that A cannot find a friend.

If there are $m$ ordinary people and $k$ honest people, denote the honest people as $A$ and the ordinary people as $B$.
(1) If $m=2 a$ is even, $k=a^{2}+1$, then take the seating arrangement
$B \cdots B, A \cdots A, B A \cdots A, B A \cdots A, \cdots, B A \cdots A$, where there are $a+1$ commas, and each comma is followed by $a+1$ people, all of whose answers are $1 \cdots 10$, so A cannot guarantee finding a friend.
(2) If $m=2 a+1$ is odd, $k=a^{2}+a+1$, then take the seating arrangement
$B \cdots B, A \cdots A, B A \cdots A, B A \cdots A, B A \cdots A$, where there are $a+1$ commas, and each comma is followed by $a+2$ people, all of whose answers are $1 \cdots 10$, so A cannot guarantee finding a friend.