Candy Delight!

Two of the children can be chosen in $\binom{20}{2}$ ways. For each of these combinations, the $20$ candies can be distributed among the remaining $18$ children, (such that each of these children receives at least one candy), in $\binom{19}{17}$ ways.

(This last calculation is a 'stars and bars' problem, an explanation of which is given here .)

Without any restrictions, the number of ways of distributing the candies is $\binom{39}{20}$ , again determined using the 'stars and bars' method. Thus the probability that exactly $2$ children receive no candies is

$\dfrac{\dbinom{20}{2} * \dbinom{19}{17}}{\dbinom{39}{20}} = \dfrac{361}{765814049}$ .

Finally, the digit sum of $361 + 765814049 = 765814410$ is $\boxed{36}$ .