14. (12 points) Three people, A, B, and C, start from $A$ to $B$ at the same time. A and B turn back to $A$ after reaching $B$, and their speeds are reduced to half of their original speeds after turning back. A turns back first, and meets B face-to-face at $C$ after turning back, at which point C has already traveled 2010 meters; A then travels for a while and meets C face-to-face at the midpoint $D$ of $A B$; B also meets C face-to-face at $C$ after turning back. Therefore, the distance between $A$ and $B$ is $\qquad$ meters.
【Solution】Solution: Let the total distance be $S$, and the time required for A and C to meet at point $D$ be $t$,
$\because$ Since A meets C at the midpoint after turning back,
$\therefore$ A has walked $\frac{3}{2}$ of the total distance, and C has walked $\frac{1}{2}$ of the total distance,
$\because$ A's speed is halved after turning back,
$\therefore$ The time before and after A's turn is the same,
$$
\begin{array}{l}
\therefore S_{\text {A }}=\frac{3}{2} S=\frac{1}{2} t V_{\text {A }}+\frac{1}{2} t \cdot \frac{1}{2} V_{\text {A }}=\frac{3}{4} V_{\text {A }} t, \\
\because S_{\text {C }}=\frac{1}{2} S=V_{\text {C }} t, \\
\therefore \frac{3}{2} S: \frac{1}{2} S=\frac{3}{4} V_{\text {A: }} V_{\text {C }} \\
\therefore V_{\text {A }}: V_{\text {C }}=4: 1,
\end{array}
$$
$A C: B C=3: 1$
If B walks at the original speed, B can walk 1.5 times the distance of $A B$ $=2010 \times 4=8040$, then the distance of $A B$ is:
$$
8040 \div 1.5=5360 \text { (meters). }
$$

Answer: The distance between $A B$ is 5360 meters.
Therefore, the answer is: 5360.