Let's find the smallest and the largest three-digit (natural) number, where the sum of the digits is 3 times larger than the sum of the digits of the number that is 75 less.
I. solution. Let $n$ be a three-digit number in which the sum of the digits is three times the sum of the digits of the number 75 less. Let the consecutive digits of $n$ be denoted by $x, y, z$, the sum of the digits by $s$, the digits of the number $n-75$ by $x_{1}, y_{1}, z_{1}$, and the sum of the digits by $s_{1}$. According to our assumption,

$$
s=3 s_{1}
$$

We will examine what cases can arise when the subtraction of $n-75$ is performed in the usual way, digit by digit, in the decimal system. What will be important for us is whether a remainder is generated when determining the digits of a given place value, or not.

$$
\begin{array}{r}
x y z \\
-\quad 75 \\
-\quad 7 \quad \\
\hline x_{1} y_{1} z_{1}
\end{array}
$$

a) No remainder is generated anywhere, that is,

$$
z \geq 5, \quad y \geq 7
$$

then $\quad x_{1}=x, \quad y_{1}=y-7, \quad z_{1}=z-5, \quad$ so

$$
s=s_{1}+12
$$

b) There is a remainder at the units place, but not at the tens place, that is,

then $\quad x_{1}=x, \quad y_{1}=y-8, \quad z_{1}=z+5, \quad$ so

$$
s=s_{1}+3
$$

c) There is no remainder at the units place, but there is at the tens place, that is,

then $\quad x_{1}=x-1, \quad y_{1}=y+3, \quad z_{1}=z-5, \quad$ so

$$
s=s_{1}+3
$$

d) There is a remainder at both the units and tens places, that is,

then $\quad x_{1}=x-1, \quad y_{1}=y+2, \quad z_{1}=z+5, \quad$ so

$$
s=s_{1}-6
$$

However, in our problem, according to (1) $s-s_{1}=2 s_{1}$, so the difference $s-s_{1}$ is positive and even, thus only case a) is possible, since in cases b) and c) this difference is odd, and in case d) it is negative. Solving the system of equations (1) and (3a) for the unknowns $s, s_{1}$, we find that

$$
s=x+y+z=18
$$

and $s_{1}=6$. If, however, the digits of the number $n$ satisfy (4) and (2a), then according to (3a) $s_{1}=s-12=6$, so (1) is also satisfied, thus the necessary and sufficient condition for the number $n$ to have the desired property is that (2a) and (4) are satisfied.

According to the problem, we need to determine the smallest and largest among these numbers. The smallest number is the one whose digits are the smallest, so its first digit is 1 and the sum of the other two is 17, hence the smallest number with the desired property is 189.

Since according to (2a) the smallest values of $z, y$ also result in their sum being 12, the largest possible value of $x$ is 6 - so the largest number with the required property is 675.

János Lengyel (Budapest, I. István Gymn., II. o. t.)

II. solution. Let the two sought numbers be $n$ and $N$. Both are divisible by 3, because the sum of their digits is three times an integer - namely, the three times the sum of the digits of the numbers $n-75$ and $N-75$.

Thus, $n-75$ and $N-75$ are also divisible by 3, because 75 is also divisible by it, and if both terms of a difference are divisible by a number, then the difference itself is also divisible by it.

Accordingly, the sum of the digits of $n-75$ and $N-75$ is of the form $3 k$ and $3 K$ (where $k, K$ are natural numbers), so the sum of the digits of $n$ and $N$ is of the form $9 k$ and $9 K$, hence $n$ and $N$ are also divisible by 9.

Knowing this, the sum of the digits of $n$ and $N$ can only be 9 and 18. Indeed, the sum of the digits of a three-digit number is at most 27 - only in the case of 999 - but this cannot be the value of $N$, otherwise the sum of the digits of $N-75$ would be 9, but 999 - 75 is not divisible by 9. Reversing our finding for $n-75$ and $N-75$, the sum of their digits can only be 3 or 6.

We will show from the other side that the sum of the digits of $n-75$ and $N-75$ cannot be 3. Indeed, on the one hand, these numbers can be written as:

$$
n-75=9 m-75=9(m-9)+6, \quad N-75=9(M-9)+6
$$

Thus, they give a remainder of 6 when divided by 9. On the other hand, in general, the (smallest non-negative) remainder of any integer when divided by 9 is the same as if we divide the sum of the digits of the number (in decimal notation) by 9. Indeed, restricting ourselves to at most three-digit numbers, a number written with the digits $A, B, C$ can be written as

$$
100 A+10 B+C=9(11 A+B)+(A+B+C)
$$

so the remainder of the division $(100 A+10 B+C): 9$ and $(A+B+C): 9$ is the same.

Accordingly, the sum of the digits of $n-75$ and $N-75$ gives a remainder of 6 when divided by 9, and combining this with the above, the sum itself is 6. From this, it follows that the sum of the digits of $n$ and $N$ is 18.

The smallest value of $n$ that fits, 189, immediately satisfies, because then $189-75=114$, the sum of the digits is 6.

The largest value of $N-75$ that fits, 600, also satisfies, because then $N=675$, the sum of the digits is 18.