Since the remainder is same in each case, hence we will use the following formula
H.C.F (x, y, z) = H.C.F of (x -y), (y- z), ( z-x)
N = H.C.F. of (6905 - 4665), (4665 - 1305), and (6905 - 1305)
    = H.C.F. of 2240, 3360 and 5600 

HCF of 5600 and 3360:

$5600=3360\times 1+2240$
$3360=2240\times 1+1120$
$2240=1120\times 2+0$
$\therefore$ HCF of 5600 and 3360 = 1120

Now, HCF of 2240 and 1120 = 1120
So, the HCF of  (3360, 2240 and 5600) = N = 1120

Sum of digits in N = (1 + 1 + 2 + 0) = 4