Four, find all such three-digit numbers. If a two-digit number itself increases by 3, then, the sum of the digits of the resulting number is equal to one third of the sum of the digits of the original three-digit number.

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Four, find all such three-digit numbers. If a two-digit number itself increases by 3, then, the sum of the digits of the resulting number is equal to one third of the sum of the digits of the original three-digit number.
Let the three-digit number $abc$ satisfy the given conditions. Then

(1) $c \geqslant 7$. In fact, if $c \leqslant b$, then the sum of the digits of $\overline{abc}+3$ will increase by 3, which does not meet the conditions.
(2) $b \neq 9$. This is because if $c \geqslant 7, b=9, a=9$, then
$$
\overline{abc}+3=1000+(c+3-10)
$$

the sum of the digits is
$$
\begin{array}{l}
1+(c+3-10)=c-6 \neq \frac{1}{3}(9+9+c) \\
\text { If } c \geq 7, \quad b=9, a<9, \text { then } \\
\overline{ab} \bar{c}+3=(a+1) \cdot 100+(c+3-10)
\end{array}
$$

the sum of the digits is
$$
(a+1)+(c+3-10)=a+c-b \text {. }
$$
Thus, $a+c-6=\frac{1}{3}(a+9+c)$ leads to $2a+2c=27$, which is impossible.
From (1) and (2), we have
$$
\begin{aligned}
a \vec{b} c+3 & =a \cdot 100+(b+1) \cdot 10 \\
& +(c+3-10),
\end{aligned}
$$

the sum of the digits is
$$
\begin{array}{l}
a+(b+1)+(c+3-10) \\
=a+b+c-6=\frac{1}{3}(a+b+c) .
\end{array}
$$

Thus, $a+b+c=9$.
Since $a \geqslant 1, c \geqslant 7$, the numbers that satisfy the conditions are:
$$
117,108,207 .
$$

Their sum is $117+108+207=432$.