Example 1: There are $n$ people registered to participate in four sports events: A, B, C, and D. It is stipulated that each person must participate in at least one event and at most two events, but events B and C cannot be registered for simultaneously. If in all different registration methods, there must be at least one method where at least 20 people register, then the minimum value of $n$ is ( ).
(A) 171
(B) 172
(C) 180
(D) 181
Let the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{T}}\right)$ represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is 1 (for example, if participating in event 甲, then $a_{\text {甲 }}=1$), otherwise, the corresponding number is 0.

Thus, each person has 9 possible ways to register for the events:
$$
\begin{array}{l}
(1,0,0,0),(0,1,0,0),(0,0,1,0), \\
(0,0,0,1),(1,1,0,0),(1,0,1,0), \\
(1,0,0,1),(0,1,0,1),(0,0,1,1) .
\end{array}
$$

Therefore, for $n$ people, there are 9 possible ways to register for the events, which can be considered as 9 drawers.
By the pigeonhole principle, when
$$
n=19 \times 9+r(r \geqslant 1)
$$

there must be at least one way in which at least 20 people have registered.
So, when $r=1$, $n$ takes the minimum value
$$
19 \times 9+1=172 \text {. }
$$

Hence, the answer is B.