jmc

In order for the equation to have to real roots, its discriminant, $b^2-4ac=(-7{)}^2-4(1)(-c)=49+4c$ must be greater than zero. So we have $$\begin{array}{rl}49+4c& >0\Rightarrow \\ 4c& >-49\Rightarrow \\ c& >\frac{-49}{4}=-\mathrm{12.25.}\end{array}$$Since $c$ must be an integer, we have $c\ge -12$.

Now we must ensure that the roots are rational. The roots are of the form $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Since $a$, $b$ and $c$ are integers, the roots are rational so long as $\sqrt{b^2-4ac}$ is rational, so we must have $b^2-4ac$ is a perfect square. Plugging in the values from our quadratic, we have $49+4c$ is a perfect square. Since $-12\le c\le 25$, we have $-48\le 4c\le 100$, so $1\le 49+4c\le 149$. There are $12$ possible squares between $1$ and $149$ inclusive, so we only need to check those $12$ squares to see if $c$ is an integer. But we can narrow this down further: the value of $49+4c$ must be odd, so it can only be the square of an odd integer. Thus the possible values for $49+4c$ are the squares of the odd numbers from $1$ to $11$. We solve:

\begin{array}{ccccc} $49+4c=1%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$4c=-48%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$c=-12$\\ $49+4c=9%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$4c=-40%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$c=-10$\\ $49+4c=25%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$4c=-24%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$c=-6$\\ $49+4c=49%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$4c=0%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$c=0$\\ $49+4c=81%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$4c=32%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$c=8$\\ $49+4c=121%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$4c=72%%DISP_1%%amp;$\Rightarrow%%DISP_1%%amp;$c=18$ \end{array}All of the values work! Their sum is $(-12)+(-10)+(-6)+0+8+18=\boxed{-2}$.