14. $A, B, C$ three people went to Teacher $D$'s house to play. Teacher $D$ gave each of them a hat and wrote a four-digit number on each hat. It is known that these three four-digit numbers are all perfect squares (for example, $4=2^{2}, 100=10^{2}, 4.100$ are all squares of some number, such numbers are called perfect squares), and the tens digit of these three four-digit numbers is 0, and the units digit is not 0. Each child can only see the numbers on others' hats. These three children are very smart and honest, and the following conversation took place:
$A$ said: “The units digit of the numbers on $B$ and $C$'s hats are the same.”
$B$ and $C$ said at the same time: “After hearing what $A$ said, I know my number.”
$A$ said: “After hearing what $B$ and $C$ said, I also know my number, and the units digit of my number is an even number.”
Find: The sum of the numbers on $A, B, C$'s hats.
【Answer】 14612

【Solution】Assume $\overline{c b 0 a}=(\overline{e f})^{2}=(10 e+f)^{2}=100 e^{2}+20 e f+f^{2}(a \neq 0 \Rightarrow f \neq 0)$, taking the modulo 100 on both sides, we get $20 e f+f^{2} \equiv a(\bmod 100)$, which means the tens digit of $20 e f+f^{2}$ is 0. Clearly, the tens digit of $20 e f$ is definitely even, so the tens digit of $f^{2}$ must also be even, and the values of $f$ that satisfy this condition are $1, 2, 3, 5, 7, 8, 9$.
(1) When $f=1, 2, 3$, to make the tens digit of $20 e f+f^{2}$ 0, then $e=5$. In this case, the three numbers are $51^{2}=2501, 52^{2}=2704, 53^{2}=2809$;
(2) When $f=5$, $20 e f+f^{2}=100 e+25$, the tens digit cannot be 0;
(3) When $f=7$, $20 e f+f^{2}=140 e+49$, to make the tens digit 0, then $e=4$ or 9, in this case, the numbers that satisfy the condition are $47^{2}=2209$ or $97^{2}=9409$;
(4) When $f=8$, $20 e f+f^{2}=160 e+64$, to make the tens digit 0, then $e=4$ or 9, in this case, the numbers that satisfy the condition are $48^{2}=2304$ or $98^{2}=9604$;
(5) When $f=9$, $20 e f+f^{2}=180 e+81$, to make the tens digit 0, then $e=4$ or 9, in this case, the numbers that satisfy the condition are $49^{2}=2401$ or $99^{2}=9801$.

Thus, we have 9 numbers that satisfy the condition, which can be grouped by the last digit into three groups: $(2209,2809,9409), (2704,2304,9604), (2501,2401,9801)$. According to the second statement by $B$ and $C$, we know that the last digits of the numbers on the hats of $A, B, C$ are the same. According to the third statement by $A$, we know that these three numbers are $(2704,2304,9604)$, so the sum of these three numbers is $2704+2304+9604=14612$.