N = 24^{3} + 25^{3} + 26^{3} + 27^{3}
Unit digits of N = 4 + 5 + 6 + 3 = 18
N is divisible by 2
We know that, a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})
⇒ (a + b) always divides a^{3} + b^{3}
Therefore, 24^{3} + 27^{3} is divisible by 51 (24 + 27)
Also, 25^{3} + 26^{3} is divisible by 51 (25 + 26)
The given divisor 102 can be expressed as 51 * 2
Hence, N is completely divisible by 102
Remainder = 0
Correct Answer: Choice D