Let N be the greatest natural number that will divide 13511, 13903 and 14589 leaving same remainder in each case. The sum of digits of N is

$$N$$ divides each of 13511, 13903, 14589 leaving remainder $$r$$ in each case.

=> $$13511=aN+r$$ -----------(i)
$$13903=bN+r$$ --------------(ii)
$$14589=cN+r$$ --------------(iii)

Subtracting above equations, we get : $$392=(b-a)N$$
$$686=(c-b)N$$
$$1078=(c-a)N$$

Thus, $$N$$ must be a factor of all three of these numbers. Thus, $$N=$$ H.C.F. (392,686,1078) = 98

$$\therefore$$ sum of digits of $$N$$ = $$9+8=17$$

=> Ans - (D)