Does such a number even exist?

The two smallest numbers having sum of digits $27$ are $999$ and $1998$ .

Since $999$ is an odd no, it can't be $2N$ . And If we assume $2N =1998$ then $N=999$ whose sum of digits are not equal to $18.$

$Thus,\ required \ no.\ is \ of \ 4 \ digits.$

So, let $N = abcd$ $\Rightarrow a+b+c+d = 18$

We are searching for smallest no., So let's assume that $a,b,c,d<5.$ then sum of digits of $2N = 2*(a+b+c+d) = 36 \neq 27$

$Thus \ our\ assumption\ is \ wrong.$

For the next smallest possibility, $a,b,c<5 \ and \ d \geq 5$

Then, sum of digits of $2N=(2a)+(2b )+(2c+1)+(2d -10) = 2*(a+b+c+d) -9 = 27$

$Thus \ our \ assumption\ is\ right\ and \ such\ a \ no. \ exists.$

So, now we know that $a,b,c<5 \ and \ d \geq 5$

For smallest possibility, $Let \ a=1$

and we know that $b+c+d =17 ,$

Thus we can easily conclude that only possibility with such restriction is $1449$ .

The smallest integer with digit sum of $27$ is $999$ , therefore, $2N$ must at least have $4$ digits and it must be even. The smallest $4$ -digit even integer with digit sum of $27$ is $1998$ but half of it is $999$ , which has digit sum of $27 \le 18$ . The next smallest $4$ -digit even integer with digit sum of $27$ is $2898$ and half of it is $1449$ , which has digit sum of $18$ . Therefore, $N = \boxed{1449}$ .

A way to do this generally would be first to observe how the digit sum is effected by doubling digits:

Let $P(N)$ be the digit sum of $N$ .

$\begin{matrix} N&0&1&2&3&4&5&6&7&8&9\\ 2N&0&2&4&6&8&10&12&14&16&18\\ \Delta P(N) = P(2N) - P(N) &+0&+1&+2&+3&+4&-4&-3&-2&-1&+0 \end{matrix}$

We want $\Delta P(N) = 9$ , we can do this with $3$ digits, but the digits will all be below 5 and $P(N) \leq 12 < 18$ . So we must use $4$ or more digits.

For $4$ digits, if our first digit is $1$ , we need the other $3$ to have $\Delta P(N) = 8$ with $P(N) = 17$ . This is only possible with $4,4,9$ . So our smallest number is $1449$ .

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Next few smallest would be $1494, 1944$ and the few after that would have first digit $2$ : The next $3$ digits will have $\Delta P(N) = 7$ , $P(N) = 16$ . Possible with $3,4,9$ and $4,4,8$ . So next comes: $2349, 2394, 2439, 2448, 2484, 2493, 2844, 2934, 2943$ .

easy hint: lowest number: 99 to keep it as low as possible we need to break one of the 9-s into some digits so that doubling those digits will give you the desired number. Carry On... :D

Any digit 5 or more, twice the sum of will not contribute to increasing the sum of digits. However the sum has to add up to 9 to be divisible by 9. We need two group, each adding to 9, with one group that when doubled would not increase the digits sum, the other that increases the digits sum .Minimum of the first non increasing group, is 9. For any other has to be two digits. The second group digits must be less than 5. But we want the smallest number . So at the heaviest position if we have 1, it would be OK. So the only combination available is (1,4,4.) and 9. 9 at any place would be good, but for minimum number, it has to be at unit place. So also, 1 has to be at the heaviest place. Thus we get $\Large1449$ . My frst thought was 33309, before the above logic..