As N be the least positive integer and when a non-zero digit C is written after the last digit of N, the resulting number is divisible by C. $\mathrm{So},10 \mathrm{N}+\mathrm{C}$ is divisible by C

$\therefore 10 \mathrm{N}$ must be divisible by C Now, the least integer (N) which is divisible by digit 'C i.e. (1 to 9) must be L.C.M of $\{1,3,4,6,7,9\}$

$\begin{array}{l}=\mathrm{L}.\mathrm{C}.\mathrm{M}\text{ of }\{4,7,9\}\\ =252=\mathrm{N}\end{array}$

and sum of digits of number ' N is

$2+5+2=9$