Let's find three-digit numbers where the sum of the digits becomes three times smaller when these numbers are tripled.
Let the digits of the number to be found be $a, b, c$; then the numbers are of the form $100a + 10b + c$. In these numbers, the last digit $c$ can only be 7, 8, or 9, because otherwise, the sum of the digits would increase by 3 when 3 is added to the number.

First, let $b < 9$. If $c = 7$, then according to the problem,

$$
a + b + 7 = 3[a + (b + 1) + 0]
$$

which gives

$$
a + b = 2
$$

and thus

$$
a_1 = 2, b_1 = 0; \quad a_2 = 1, b_2 = 1
$$

so the numbers are 207, 117.

If $c = 8$, then

$$
a + b + 8 = 3[a + (b + 1) + 1]
$$

which gives

$$
a + b = 1
$$

and thus

$$
a = 1, b = 0
$$

so the number is 108.

If $c = 9$, then

$$
a + b + 9 = 3[a + (b + 1) + 2]
$$

which gives

$$
a + b = 0
$$

and thus there is no solution in this case.

Second, let $b = 9$. Then the conditional equations are respectively

$$
\begin{aligned}
& a + 16 = 3(a + 1) \\
& a + 17 = 3(a + 2) \\
& a + 18 = 3(a + 3)
\end{aligned}
$$

Since in these equations $a$ has fractional values, there are no solutions in this case. The sought numbers are therefore $108, 117, 207$.

(Júlia Póka, Losoncz.)

The problem was also solved by: B. Bayer, K. Burján, K. Czank, F. Faith, J. Filkorn, D. Kőnig, B. Krausz, Gy. Krisztián, A. Kürth, M. Lindtner, Gy. Lukhaub, Gy. Lupsa, G. Messik, V. Messik, Gy. Perl, G. Sasvári, J. Sasvári, J. Scharff, J. Szabó.