2. As shown in Figure 1, given that $G$ is the centroid of $\triangle A B O$. If $P Q$ passes through point $G$, and
$$
\begin{array}{l}
O A=a, O B=b, \\
O P=m a, O Q=n b,
\end{array}
$$

then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$
2.3.

From $O M=\frac{1}{2}(a+b)$, we know
$$
O G=\frac{2}{3} O M=\frac{1}{3}(a+b) .
$$

From the collinearity of points $P, G, Q$, we have $\boldsymbol{P G}=\lambda \boldsymbol{G} \boldsymbol{Q}$.
And $P G=O G-O P=\frac{1}{3}(a+b)-\dot{m} a$
$$
\begin{array}{l}
=\left(\frac{1}{3}-m\right) a+\frac{1}{3} b, \\
G Q=O Q-O G=n b-\frac{1}{3}(a+b) \\
=-\frac{1}{3} a+\left(n-\frac{1}{3}\right) b,
\end{array}
$$

Thus, $\left(\frac{1}{3}-m\right) a+\frac{1}{3} \boldsymbol{b}$
$$
=\lambda\left[-\frac{1}{3} \boldsymbol{a}+\left(n-\frac{1}{3}\right) \boldsymbol{b}\right] \text {. }
$$

Since $a, b$ are not collinear, we have
$$
\left\{\begin{array}{l}
\frac{1}{3}-m=-\frac{1}{3} \lambda, \\
\frac{1}{3}=\lambda\left(n-\frac{1}{3}\right) .
\end{array}\right.
$$

Solving these equations, we get $3 m n=m+n$.
Therefore, $\frac{1}{m}+\frac{1}{n}=3$.