Since two of the lines that are the edges of the triangle are known, their intersection must be one of the vertices of the triangle. So we have $ y=0 $ (the $ x $-axis) and $ y=\frac{2}{3}x+5 $. Solving this equation, we find $ 0=\frac{2}{3}x+5 $, or $ -5=\frac{2}{3}x $, so $ x=-\frac{15}{2} $. Thus one of the vertices of the triangle is $ \left(-\frac{15}{2},0\right) $. The other vertices lie on the line $ x=k $, so they take the form $ (k,0) $ and $ \left(k,\frac{2}{3}k+5\right) $. The area of the triangle can be expressed as $ \frac{1}{2}bh $. The height is $ \frac{2}{3}k+5 $, since the base is along the $ x $-axis, and the base is $ k-\left(-\frac{15}{2}\right)=k+\frac{15}{2} $. Thus the area is $ \frac{1}{2}\left(\frac{2}{3}k+5\right)\left(k+\frac{15}{2}\right) $. Up until this point we've mostly ignored the possibility of having a triangle below the $ x $-axis, with $ k<-\frac{15}{2} $. This is possible, but our formula for the area will still work. If $ k<-\frac{15}{2} $, then $ k+\frac{15}{2} $ will be negative. But the line $ y=\frac{2}{3}x+5 $ will be below the $ x $-axis so the value $ \frac{2}{3}k+5 $ will be negative as well. Half their product, the area, will thus be positive as desired. So we have \begin{align*} \frac{1}{2}\left(\frac{2}{3}k+5\right)\left(k+\frac{15}{2}\right)&<20\quad\Rightarrow\\ \left(\frac{2}{3}k+5\right)\left(k+\frac{15}{2}\right)&<40\quad\Rightarrow\\ \frac{2}{3}k^2+10k+\frac{75}{2}&<40\quad\Rightarrow\\ \frac{2}{3}k^2+10k-\frac{5}{2}&<0\quad\Rightarrow\\ 4k^2+60k-15&<0.\end{align*}We must solve this quadratic inequality. The roots of the quadratic are $$\frac{-(60)\pm\sqrt{(60)^2-4(4)(-15)}}{2(4)}=\frac{-60\pm\sqrt{3840}}{8}=-\frac{15}{2}\pm2\sqrt{15}.$$Testing, we find that the value of the quadratic is negative between the roots, or $ 4k^2+60k-15<0 $ whenever $ -\frac{15}{2}-2\sqrt{15}<k<-\frac{15}{2}+2\sqrt{15} $. The decimal approximations of the roots are $ -15.25\ldots $ and $ 0.25\ldots $, respectively, so we have $ -15.25<k<0.25 $. Since we are interested in values of $ k $ for which $ k $ is an integer, we have $ -15\le k\le 0 $. The problem asks for the sum of all integral values of $ k $, so we must sum the integers from $ -15 $ to $ 0 $. We can compute this with the formula for the sum of an arithmetic series: $ S=\frac{(-15+0)(16)}{2}=\boxed{-120} $.