Example 4 As shown in Figure 2, let $M$ be the centroid of $\triangle A B C$, and a line through $M$ intersects sides $A B$ and $A C$ at points $P$ and $Q$, respectively, and
$$
\frac{A P}{P B}=m, \frac{A Q}{Q C}=n \text {. }
$$

Then $\frac{1}{m}+\frac{1}{n}=$
Solution 1 As shown in Figure 2, draw lines through points $C$ and $B$ parallel to $PQ$ intersecting $AD$ or its extension at points $E$ and $F$. Then,
$$
\begin{array}{l}
\frac{1}{m}=\frac{P B}{A P}=\frac{F M}{A M}, \\
\frac{1}{n}=\frac{Q C}{A Q}=\frac{E M}{A M} .
\end{array}
$$

Since $D$ is the midpoint of side $BC$, and $BF \parallel CE$, it follows that $DE = DF$.
Thus, $FM + EM = 2DM$.
Furthermore, since $M$ is the centroid of $\triangle ABC$, we have $AM = 2DM$. Therefore, $\frac{1}{m} + \frac{1}{n} = \frac{FM + EM}{AM} = \frac{2MD}{AM} = 1$.
Solution 2 As shown in Figure 3, consider a special case where $PQ \parallel BC$.
Then,
$$
\begin{array}{l}
\frac{A P}{P B}=\frac{A Q}{Q C} \\
=\frac{A M}{M D}=2 .
\end{array}
$$

Thus,
$$
\ldots \frac{1}{m}+\frac{1}{n}=1 \text {. }
$$