Three, (16 points) Remove the middle digit of a three-digit number $\overline{a b c}$, making it a two-digit number $\overline{a c}$, and satisfy $\overline{a b c}=$ $9 \overline{a c}+4 \bar{c}$ (for example, $155=9 \times 15+4 \times 5$). Try to find all such three-digit numbers.
Three, since $\overline{a b c}=100 a+10 b+c, \overline{a c}=10 a+c$, from the problem we get $100 a+10 b+c=9(10 a+c)+4 c$. Simplifying, we get
$$
5(a+b)=6 c \text {. }
$$

Here $0 \leqslant a, b, c \leqslant 9$, and $a \neq 0$.
Since 5 is a prime number, we have $\left\{\begin{array}{l}c=5, \\ a+b=6 \text {. }\end{array}\right.$
Thus, $\left\{\begin{array}{l}a=1,2,3,4,5,6, \\ b=5,4,3,2,1,0 .\end{array}\right.$
Therefore, $\overline{a b c}=155,245,335,425,515,605$.