14. (16 points) As shown in Figure 4, $A$ and $B$ are the common vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. $P$ and $Q$ are moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})(\lambda \in \mathbf{R},|\lambda|>1) \text {. }
$$

Let $k_{1} 、 k_{2} 、 k_{3} 、 k_{4}$, then $k_{1}+k_{2}+k_{3}+k_{4}$ is a constant.
Prove: (1) $O, P, Q$ are collinear;
(2) If the slopes of the lines $AP, BP, AQ, BQ$ are respectively
14. (1) Note that,
$\overrightarrow{A P}+\overrightarrow{B P}=2 \overrightarrow{O P}, \overrightarrow{A Q}+\overrightarrow{B Q}=2 \overrightarrow{O Q}$.
Also, $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})$, hence $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$.
Therefore, $O, P, Q$ are collinear.
(2) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}^{2}-a^{2}=\frac{a^{2}}{b^{2}} y_{1}^{2} \text {. }
$$

Thus, $k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}$
$$
=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}} \text {. }
$$

Similarly, $k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}$.
From (1), we know that $O, P, Q$ are collinear, i.e., $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$.
Therefore, from equations (1) and (2), we get
$$
\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}} \\
=\frac{2 b^{2}}{a^{2}}\left(\frac{x_{1}}{y_{1}}-\frac{x_{2}}{y_{2}}\right)=0 .
\end{array}
$$