jmc

Let the page numbers be $n$ and $n+1.$ Then, the problem can be modeled by the equation $n(n+1)=18360.$ We can rewrite the equation as $n^2+n-18360=0.$

Now using the quadratic formula, we find that $$n=\frac{-1\pm \sqrt{1+4\cdot 18360}}{2}.$$ So, $n=135.$ Hence, $n+(n+1)=\boxed{271}.$

This equation can be factored as well, but that would not save much time. The best way to solve this quickly would be to notice that $18360$ falls between $135^2=18225$ and $136^2=18496,$ so since we know that $n$ is an integer, we can guess that $n=135.$ Plugging it back into the equation, we see that it works, so $n+(n+1)=\boxed{271}.$