14. A non-zero natural number $\mathrm{n}$, is both the sum of 2010 natural numbers with the same digit sum, the sum of 2012 natural numbers with the same digit sum, and the sum of 2013 natural numbers with the same digit sum. What is the smallest value of $m$?
Analysis: Number Theory. According to the problem, we have $n=2010 \mathrm{~A}=2012 \mathrm{~B}=2013 \mathrm{C}$. The numbers that can connect the sum of digits and the number itself are those divisible by 3 or 9. Understand a conclusion: to determine if a number is divisible by 3 or 9, split the number into several parts by its digits. If the sum of these parts is divisible by 3 or 9, then the number is divisible by 3 or 9. For example, to determine if $12346789101112 \cdots 2013$ is divisible by 9, we only need to check if the sum $1+2+\cdots+2013$ is divisible by 9.
Clearly, 2010 and 2013 are multiples of 3, so $\mathrm{n}$ is a multiple of 3, and $2012 \mathrm{~B}$ is a multiple of 3. According to the non-zero natural numbers, $\mathrm{B}$ is at least 3, so $\mathrm{n}$ is at least 6036.
Verification: $x+y=2010 \quad 3 x+12 y=6036, x=2008, y=2$ (the sum of digits can also be 2)
$c+d=201310 c+d=6036 \quad c=447, d=1566$ (the sum of digits equal to 2 and 3 is not possible)
$$
6036=2012 \times 3=2008 \times 3+12 \times 2=10 \times 447+1566 \times 1
$$

The minimum value of the total $n$ is 6036.