$ 2449999999999$.

How do you get it?

Minimal number, when it had minimal number od digits or maximum digits equal 9. Because $ S(2N)-S(N)=10$, the sum of digits <5 is 10 and 100=10+10*9. Minimal case when $ 2449...9 (ten \ 9)$

Call the number $ \overline{abcdefg.....}$ Consider multiplying a number by 2 by hand. For a digit less than 5, there won't be a carry, so the resulting digit will be twice the original. Otherwise, you will carry the one, so the resulting digit will be twice it minus 10, but because of the carry, you will add one to the next number. So for each digit n, the amount contributed to the sum of the digits of twice the number is either 2n or 2n-9. So if we subtract these from the original digit to get the difference in the sums, each digit contributes either n or n-9 and the difference in the digit sums is 10. So we have a+b+c+d...=100 and {a or a-9}+{b or b-9}+{c or c-9}...=10. Hence, 10 of the digits must contribute n-9, so there must be 10 carries, so since the digits add up to 100, the least is 2449999999999.

we only have to consider how much times the carries occur,and can easily get the answer $249999999999$. (if there is no carry,then the digit sum of $2N$ will be 200;since the digit sum will decrease by $9$ after we increase a craay,hence $10$ carries must occurs).