6. (10 points) From the natural numbers $1,2,3, \cdots, 2015,2016$, arbitrarily select $n$ different numbers, requiring that among these $n$ different numbers, it is always possible to find 5 numbers whose digit sums are equal. Then the minimum value of $n$ is ( )
A. 109
B. 110
C. 111
D. 112
【Analysis】First, determine that the problem requires the sum of the digits of each number. The sum of the digits of 120 is 3. Therefore, the largest sum of the digits is 28, which is for 1999, and the smallest sum is 1, which occurs for the number 1. There are at most 4 numbers that can be selected. Selecting one more will meet the condition.

【Solution】According to the problem:
In the range $1-2019$, the largest sum of the digits is 28, which is for the number 1999.
The smallest sum of the digits is 1, which occurs for the numbers $1, 10, 100, 1000$, a total of four numbers.
The sum of the digits being 27 occurs for $999, 1899, 1998, 1989$, a total of four numbers.
The sum of the digits being $2-26$ all exceed 5 numbers.
Therefore, we just need to pick 4 numbers from the sums of the digits being $2-26$, and include the sums of the digits being $1, 27, 28$. Adding one more will make it 5 numbers that meet the condition.
$$
27 \times 4 + 1 + 1 = 110 \text{.}
$$

Therefore, the answer is: $B$.