jmc

Call the point where the the runner touches the wall $C$. Reflect $B$ across the wall to $B^{\prime }$. Since $CB=C{B}^{\prime }$, minimizing $AC+CB$ is equivalent to minimizing $AC+C{B}^{\prime }$. The wall is between $A$ and $B^{\prime }$, so we may choose $C$ on line segment $A{B}^{\prime }$. This choice minimizes $AC+C{B}^{\prime }$, because the shortest distance between two points is a straight line. By the Pythagorean theorem, $A{B}^{\prime }=\sqrt{1200^2+(300+500{)}^2}=400\sqrt{13}$ meters, which to the nearest meter is $\boxed{1442}$ meters.