Jane can walk any distance in half the time it takes Hector to walk the same distance.  They set off in opposite directions around the outside of the 18-block area as shown.  When they meet for the first time, they will be closest to

$\text{(A)}\ A \qquad \text{(B)}\ B \qquad \text{(C)}\ C \qquad \text{(D)}\ D \qquad \text{(E)}\ E$
Counting around, when Jane walks $12$ steps, she will be at $D$.  When Hector walks $6$ steps, he will also be at $D$.  Since Jane has walked twice as many steps as Hector, they will reach this spot at the same time.  Thus, the answer is $\boxed{D}$.