Which are the three-digit natural numbers, where the sum of the digits decreases to a third of its original value when 3 is added to the number?
The last digit of the sought $N$ can only be a digit greater than 6, because otherwise the third digit of the $N^{\prime}$, which is three times larger, would be at most 9, and thus larger than the third digit of $N$, with the first two digits matching those of $N$, so the sum of the digits of $N^{\prime}$ could not be smaller than the sum of the digits of $N$.

Thus, the last digit of $N$ is 7, 8, or 9, and as a result, the last digit of $N^{\{\prime\}}$ deviates by $+3-10=-7$, or 7 less, due to the carry-over of 1 to the tens place of $N$. Depending on whether this digit is less than 9 or equal to 9, two cases arise; the hundreds place changes only in the latter case.

If the tens place digit of $N$ is less than 9, then the tens place digit of $N^{\prime}$ is 1 greater, so the sum of the digits of $N^{\prime}$ is only $7-1=6$ less than that of $N$. This decrease is one-third of the sum of the digits of $N$, so the sum of the digits of $N$ is 9. - If the last digit of $N$ is already 7, then the sum of the first two digits is 2, from which 207, 117, and 027 are obtained for $N$, but the latter is not a proper three-digit number. Similarly, we get the numbers 108 and 018 for the last digit being 8, and 009 for the last digit being 9.

If the tens place digit of $N$ is 9, then the tens place digit of $N^{\prime}$ is 0 from 10, or 9 less, and the hundreds place digit is 1 greater - unless the hundreds place digit of $N$ is also 9. If the hundreds place digit of $N$ is less than 9, then the hundreds place digit of $N^{\prime}$ is 1 greater, so the sum of the digits of $N^{\prime}$ decreases by $7+9=16$ and increases by 1, resulting in a net decrease of 15. However, this cannot be one-third of an integer, so there is no solution to the problem in this case. - If, finally, the hundreds place digit of $N$ is also 9, then the sum of the digits of $N^{\prime}$ would decrease by $7+9+9=25$ and increase by 1 in the thousands place, resulting in a decrease of 24. - This would mean the sum of the digits is 36, which is impossible, because the sum of the digits of a three-digit number can be at most $3 \cdot 9=27$, - so there is no solution in this case either.

Therefore, the only solutions to our problem are the numbers 207, 117, and 108. All three are valid, because in the numbers 210, 120, and 111, which are three times larger, the sum of the digits is 3, one-third of 9.

Remarks. 1. Many contestants found the above numbers by trial and error. However, they did not prove that only these three solutions exist. - Some stated without proper justification that the sum of the digits can only be 9, or that the last digit can only be 7, 8, or 9. None of these can be considered a complete solution.

2. The following line of thought is more elegant but requires much more work to reach the goal. Since the sum $s^{\prime}$ of the digits of $N^{\prime}$ is one-third of the sum of the digits of $N$, the sum of the digits of $N$ is: $s=3s^{\prime}$, divisible by 3, and therefore $N$ itself is also divisible by 3. Thus, $N^{\prime}=N+3$ is also divisible by 3, so this is true for $s^{\prime}$ as well: $s^{\prime}=3k$, and thus $s=9k$. The sum of the digits of a three-digit number is at most 27, so $s=9$, or 18, or 27. It is easy to show that 999, the only three-digit number with a digit sum of 27, is not a solution, but examining the possibility of $s=18$ is tedious. From the assumption $s=9$ - similarly to the above - the three solutions easily follow.