Which is the three-digit (integer) number that, when increased or decreased by the sum of its digits, results in a number consisting of the same digit repeated?
The sought number is the arithmetic mean of the numbers obtained by increasing and decreasing it, which are all numbers with identical digits, because the performed increase and decrease are equal. On the other hand, the difference between the resulting numbers is twice the sum of the digits.

The resulting numbers can be numbers of the form $\overline{a a a}=111 \cdot a$. It is conceivable that a four-digit number could result from adding the sum of the digits, and a two-digit number from subtraction. However, it is not possible for both resulting numbers to be three-digit, because the difference between two such numbers is at least 111, while twice the sum of the digits can be at most $2(9+9+9)=54$. This also implies that a four-digit number cannot result from the increase, because the smallest such number is 1111, and this is 112 more than the largest possible three-digit number, 999.

According to this, the number obtained by increasing can only be 111, otherwise twice the sum of the digits would again exceed 54, and the number obtained by decreasing must be two-digit. Since the original number is at least 100, the sum of the digits cannot be more than $111-100=11$, so the number obtained by decreasing cannot be less than $100-11=89$, thus it can only be 99.

Given all of this, the sought number can only be $(99+111) / 2=105$. This indeed fits, because the sum of its digits is 6 and $105+6=111,105-6=99$.