Example 3 Suppose the positive integer $n$ can be expressed as the sum of the reciprocals of four distinct positive integers. Then the number of such $n$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
(2011, Fudan University Independent Recruitment Examination)
Let $n=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$, and $a<b<c<d$ $\left(a, b, c, d \in \mathbf{Z}_{+}\right)$.
If $a \geqslant 3$, then $b \geqslant 4, c \geqslant 5, d \geqslant 6$.
Thus, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$
$$
\leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{19}{20}<1 \text {. }
$$

Therefore, $a=1$ or 2.
When $a=1$,
$$
1<n \leqslant 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}<3 .
$$

Thus, $n$ can only be 2, and
$$
2=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \text {. }
$$

When $a=2$,
$$
n \leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{77}{60}<2 \text {. }
$$

Thus, $n$ can only be 1, and
$$
1=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12} \text {. }
$$

In summary, there are 2 values of $n$ that satisfy the conditions, so the answer is (B).