jmc

Let our integer be $x$. Then we have that $x^2=182+x$, or $x^2-x-182=0$. The sum of the roots of this equation is just $-(-1)=\boxed{1}$. Note that we are given that one solution is an integer, and so the other one must be as well since they add to 1.

Note that we can factor $x^2-x-182=0$ as $(x-14)(x+13)=0$. So the integers that work are 14 and $-13$, and their sum is $14+(-13)=1,$ as expected.