14. As shown in the figure, in trapezoid $\mathrm{ABCD}$, $CD = 2AB$, points $E, F$ are the midpoints of $AD, AB$ respectively. If the area of triangle $CDG$ minus the area of quadrilateral AEGF equals $\frac{k}{24}$ square centimeters (where $\mathbf{k}$ is a positive integer), to make the area of trapezoid $\mathrm{ABCD}$ a positive integer, then the minimum value of $k$ is $\qquad$
【Analysis】Let $A B=2 a, C D=2 b$, and the height be $h$.
$$
\begin{array}{l}
\mathrm{S}_{\text {trapezoid }}(-\mathrm{AB}+C D) \times \mathrm{h} \div 2=(2 a+4 a) \times \mathrm{h} \div 2=3 a h \\
\because S_{\triangle C D G}-S_{A E G F}=\frac{\mathrm{k}}{24} \\
\therefore\left(S_{\triangle C D G}+S_{\triangle D E G}\right)-\left(S_{A E G F}+S_{\triangle D E G}\right)=\frac{\mathrm{k}}{24} \\
\therefore S_{\triangle C D E}-S_{\triangle A D F}=\frac{\mathrm{k}}{24} \\
\therefore \frac{1}{2} \times 4 a \times \frac{1}{2} \mathrm{~h}-\frac{1}{2} \times a \times \mathrm{h}=\frac{\mathrm{k}}{24} \\
\therefore \frac{1}{2} a \mathrm{~h}=\frac{\mathrm{k}}{24} \\
a \mathrm{~h}=\frac{\mathrm{k}}{12} \\
\therefore S_{\text {trapezoid }}=3 a h=3 \times \frac{\mathrm{k}}{12}=\frac{\mathrm{k}}{4} \text { (an integer) } \\
\therefore \mathrm{k}_{\min }=4
\end{array}
$$