11. (10 points) A three-digit number $N$, when subtracted by 3, added by 4, divided by 5, and multiplied by 6, results in four integers. It is known that the sum of the digits of these four numbers are exactly 4 consecutive natural numbers. How many three-digit numbers $N$ satisfy this condition?
A. 8
B. 6
C. 4
D. 2
【Solution】Solution: Considering there will definitely be carrying and borrowing, let the sum of the original number's digits be $a$. Then, $-3, +4$ cannot have a difference of 7, otherwise it cannot form 4 consecutive natural numbers. $\div 5$ indicates the last digit is 0 or 5. When the last digit is 5, $-3, +4$ do not involve carrying or borrowing; when the last digit is 0, -3 involves borrowing, which fits,

So -3 is equivalent to the sum of the digits being 6 more, $a+6$; +4 is equivalent to the sum of the digits being 4 more, $a+4$; $\div 5$ is equivalent to the sum of the digits being multiplied by 2, $a \times 2$; $a \times 2, a+2, a+4$ are consecutive, $a \times 2$ is one of $a+7, a+5, a+3$,

By categorizing and discussing, we get $a \times 2 = a+5$ holds true, so $a=5$, the sum of the digits is 5, and the numbers ending in 0 are: 500 (discard), 410, 320, 230, 140, a total of 4.

Therefore, the answer is: $C$.