Let's find a three-digit number where the sum of the digits is equal to the difference between the number formed by the first two digits and the number formed by the last two digits.
Let $x, y, z$ represent the three digits of the sought number. Then, according to the problem:

$$
x+y+z=(10 x+y)-(10 y+z)
$$

or

$$
9 x=10 y+2 z
$$

from which

$$
x=y+\frac{y+2 z}{9}
$$

For $x$ to be an integer, $y+2 z$ must be a multiple of 9, and since $0 \leq y \leq 9$ and $0 \leq z \leq 9$, $y+2 z$ can be at most 27. This latter value is excluded, because otherwise $x$ would be greater than 9.

Since 9 and 18 are the only multiples of 9 that fall between 0 and 27, we have two assumptions:

$1^{0} \quad y+2 z=9 \quad$ and $\quad x=y+1$.

From these two equalities, it follows that $y$ can only be an odd number less than 8, yielding the following four solutions:

$$
\begin{array}{llll}
y=1, & 3, & 5, & 7 ; \\
z=4, & 3, & 2, & 1 ; \\
x=2, & 4, & 6, & 8
\end{array}
$$

$2^{0} \quad y+2 z=18 \quad$ and $\quad x=y+2$.

These two equalities require that $y$ be an even number less than 8, yielding the following four new solutions:

$$
\begin{array}{llll}
y=0, & 2, & 4, & 6 ; \\
z=9, & 8, & 7, & 6 ; \\
x=2, & 4, & 6, & 8
\end{array}
$$

Thus, the following 8 numbers satisfy the problem:

$$
\begin{array}{llll}
209, & 428, & 647, & 866 \\
214, & 433, & 652, & 871
\end{array}
$$

(Leó Goldberger, realist. VII. grade. Pécs).

The problem was also solved by: Berháth Friedmann, S. A. Ujhely; János Imre, Jr., Nyíregyháza; Elemér Meitner, Budapest; Imre Szentpétery, Losoncz; Aladár Visnya and Lipót Weisz, Pécs.