jmc

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 $$\begin{array}{rl}\frac{1}{2}\left(\frac{2}{3}k+5\right)\left(k+\frac{15}{2}\right)& <20\Rightarrow \\ \left(\frac{2}{3}k+5\right)\left(k+\frac{15}{2}\right)& <40\Rightarrow \\ \frac{2}{3}{k}^2+10k+\frac{75}{2}& <40\Rightarrow \\ \frac{2}{3}{k}^2+10k-\frac{5}{2}& <0\Rightarrow \\ 4k^2+60k-15& <0.\end{array}$$We must solve this quadratic inequality. The roots of the quadratic are $$\frac{-(60)\pm \sqrt{(60{)}^2-4(4)(-15)}}{2(4)}=\frac{-60\pm \sqrt{3840}}{8}=-\frac{15}{2}\pm 2\sqrt{15}.$$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 $-15.25\dots$ and $0.25\dots$, 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: $S=\frac{(-15+0)(16)}{2}=\boxed{-120}$.