Example 3 A philanthropist recruits members for his club in the following way: each member can introduce two others to join, where these two are not introduced by anyone else; and each new member can also introduce two others to join. For a member $A$, the members introduced by $A$ and the members introduced by those introduced by $A$ are collectively referred to as $A$'s "downlines." If each of the two members $B_{1}, B_{2}$ introduced by a member $A$ has at least 200 downlines, then member $A$ will receive a dinner voucher from the philanthropist as a reward at the end of the year.

Determine: If a total of 2012 people have become members, then at most how many people can enjoy the philanthropist's dinner at the end of the year?
Solve using reverse thinking.
For this, let $200=k$, and denote the minimum number of members when exactly $r$ people receive coupons as $f(r)$. Members who join without being introduced (or without a superior) are called "bosses", and the members they introduce are called "followers". Each boss along with all their followers and subordinates form a "gang"; each gang forms a tree diagram.

To minimize the number of members, in any gang, a boss who cannot win a prize can actually be removed. Thus, some of the followers under their control may become bosses, forming new gangs along with the subtrees they manage.

When $r=1$, the minimum number of members is $2k+3$ (at this time, only $A$ wins, and $A, B_1, B_2$ are all members, and $B_1, B_2$ each have at least $k$ subordinates), i.e., $f(1)=2k+3$.

When $r=2$, the winners are $A$ and one of $B_1, B_2$ (let it be $B_1$), and according to the above discussion, the branch containing $B_1$ (including $B_1$) has at least $2k+3$ people. Since $A$ is a winner, $B_2$ must have at least $k$ subordinates. Therefore, the number of members is at least
$$
1+(2k+3)+(k+1)=3k+5,
$$
i.e., $f(2)=3k+5$.
When $r=3$, there are two scenarios for the winners.
(1) $A, B_1, B_2$;
(2) $A, B_1$, and one of $B_1$'s subordinates (let it be $C_1$).
For (1), according to the above discussion, the branch containing $B_1$ (including $B_1$) has at least $2k+3$ people; the branch containing $B_2$ (including $B_2$) also has at least $2k+3$ people. Therefore, including $A$, the number of members is at least $4k+7$.

For (2), $B_1$'s branch has two winners, and this branch has at least $3k+5$ people; since $A$ is a winner, $B_2$ must have at least $k$ subordinates. Therefore, the number of members is at least
$$
1+(3k+5)+(k+1)=4k+7.
$$

Thus, both scenarios result in $f(3)=4k+7$.
$$
\text{By } f(2)-f(1)=k+2, f(3)-f(2)=k+2,
$$

we can guess that for each positive integer $n$,
$$
\begin{array}{l}
f(r+1)-f(r)=k+2. \\
\text{Thus, } f(r)=r(k+2)+(k+1).
\end{array}
$$

It is sufficient to prove equation (1) for the case of only one gang.
In fact, if there are $s$ gangs, each gang has $n_1, n_2, \cdots, n_s$ people, and the number of winners are $r_1, r_2, \cdots, r_s$, then
$$
\begin{array}{l}
\text{Total number of members } n=\sum_{i=1}^{1} n_{i}, \\
\text{Total number of winners } r=\sum_{i=1}^{1} r_{i}. \\
\text{Thus, } n=\sum_{i=1}^{s} n_{i} \geqslant \sum_{i=1}^{\prime} f\left(r_{i}\right) \\
=\sum_{i=1}^{s}\left[r_{i}(k+2)+(k+1)\right] \\
=(k+2) \sum_{i=1}^{s} r_{i}+s(k+1) \\
=r(k+2)+s(k+1) \\
\geqslant r(k+2)+(k+1).
\end{array}
$$

Next, we prove equation (1) using mathematical induction.
When $r=1,2,3$, it has already been established.
Assume equation (1) holds for fewer than $r$ winners.
When there are $r$ winners and the number of members in this gang has reached the minimum, the boss $A$ must be a winner. Let the two direct followers of boss $A$ be $B, C$. If the number of people in $B$'s sub-gang (including $B$ and all their followers) is $n_B$, and the number of winners among them is $r_B$, similarly define $n_C, r_c$.

According to the induction hypothesis, if $r_B>0$ (i.e., $B$'s sub-gang has winners), then the number of members in this sub-gang is
$$
n_B \geqslant r_B(k+2)+(k+1).
$$

If $r_B=0$, since $A$ is a winner, $B$ must have introduced at least $k$ people, including $B$ themselves, $B$'s sub-gang has at least $k+1$ members. Thus,
$$
n_B \geqslant k+1=r_B(k+2)+(k+1).
$$

Therefore, in any case,
$$
n_B \geqslant r_B(k+2)+(k+1).
$$

Similarly, $n_C \geqslant r_c(k+2)+(k+1)$.
Since the total number of members in $A$'s gang is $n=n_B+n_C+1$, and the total number of winners is $r=r_B+r_C+1$, we have
$$
\begin{array}{l}
n=n_B+n_C+1 \\
\geqslant\left(r_B+r_C\right)(k+2)+2(k+1)+1 \\
=r(k+2)+(k+1), \\
f(r)=r(k+2)+(k+1).
\end{array}
$$

Thus, $f(r)=r(k+2)+(k+1)$.
According to equation (1), when there are $r$ winners, the number of members $n$ satisfies $n \geqslant f(r)=r(k+2)+(k+1)$.
When $n=2012, k=200$, the above equation becomes
$$
202 r \leqslant 1811 \Rightarrow r \leqslant\left[\frac{1811}{202}\right]=8.
$$

Therefore, at most 8 people can receive the philanthropist's banquet at the end of the year.