10. Among the positive integers less than 10000, by swapping the digits in the highest and lowest positions of a number, a new number is obtained, and this new number is 1.2 times the original number. The sum of all numbers that satisfy the above condition is $\qquad$ .
【Analysis】Since the new number is 1.2 times the original number, let the original number be $N$, then $1.2 N$ is an integer, so the unit digit of $N$ must be 0 or 5. If the unit digit of $N$ is 0, then by swapping the highest and lowest digits of $N$, the resulting number will definitely be smaller, and it cannot be 1.2 times $N$, so the unit digit of $N$ must be 5.
One, there is clearly no such number among single-digit numbers.
Two, among two-digit numbers, let's assume the original number is $\overline{a 5}$, according to the problem, we have
$$
\begin{array}{l}
\overline{5 a}=1.2 \bar{a} \\
50+a=1.2(10 a+5) \\
11 a=44 \\
a=4
\end{array}
$$

So the two-digit number that meets the condition is 45
Three, among three-digit numbers, let's assume the original number is $\overline{a b 5}$, according to the problem, we have $\overline{5 b a}=1.2 \overline{a b 5}$
$$
\begin{array}{l}
500+10 b+a=1.2(100 a+10 b+5) \\
119 a+2 b=494
\end{array}
$$

Since $a, b$ can only be selected from 0 to 9, and $a$ is not 0,
there is only one solution $\left\{\begin{array}{l}a=4 \\ b=9\end{array}\right.$
So the three-digit number that meets the condition is 495
Four, among four-digit numbers, let's assume the original number is $\overline{a b c 5}$, according to the problem, we have
$$
\begin{array}{l}
\overline{5 b c a}=1.2 \overline{a b c 5} \\
5000+100 b+10 c+a=1.2(1000 a+100 b+10 c+5) \\
1199 a+20 b+2 c=4994
\end{array}
$$

Since $a, b, c$ can only be selected from 0 to 9, and $a$ is not 0,
there is only one solution $\left\{\begin{array}{l}a=4 \\ b=9 \\ c=9\end{array}\right.$
So the four-digit number that meets the condition is 4995
In summary, the sum of all numbers that meet the condition is $45+495+4995=5535$