Solution:

Let x be the required remainder. Then N - x divides each of 1305 - x, 4665 - x and 6905 - x.


HCF of (1305 - x), (4665 - x) and (6905 - x) is the same as that of (1305 - x), (4665 - x - 1305 + x) = (3360 - x) and (6905 - x - 1305 + x) = (5600 - x).
Therefore, N - x is a factor of HCF of (3360 - x), (5600 - x) and (4665 - 1305) = 3360, 5600 and 1360.
Also, N - x is a factor of HCF of (3360 - x), (5600 - x) and (6905 - 1305) = 5600.
Therefore, N - x is a factor of HCF of 3360, 5600 and 1360, which is 80.

Therefore, the required number N is of the form 80k + x where k is an integer.
Given that N is the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case.
Therefore, x is the HCF of (1305, 4665, 6905) = 85.

Hence, N = 80k + 85.

Now we need to find the sum of digits in N.
80k + 85 = 5(16k + 17) => N is of the form 5p + 4 where p = 16k + 17.

Therefore, we need to find the sum of digits in a number of the form 5p + 4.

When p = 0, N = 4, sum of digits = 4.
When p = 1, N = 9, sum of digits = 9.
When p = 2, N = 14, sum of digits = 5.
When p = 3, N = 19, sum of digits = 10.
When p = 4, N = 24, sum of digits = 6.
When p = 5, N = 29, sum of digits = 11.
When p = 6, N = 34, sum of digits = 7.
When p = 7, N = 39, sum of digits = 12.
When p = 8, N = 44, sum of digits = 8.
When p = 9, N = 49, sum of digits = 13.
When p = 10, N = 54, sum of digits = 9.

We observe that the sum of digits repeats itself after every 5 values of p.

Therefore, when p = 16k + 17, sum of digits = sum of digits when p = 2, which is 5.

Hence, the sum of digits in N is 5, which is option (a).