M 2014+2011m is divisible by 5, which requires that the last digit of m 2014+201m is 5 or 0.

Obviously, the last bit of 20 1 1 m is 1. If the last digit of m 2014 is 4 or 9, their sum can meet the requirements.

When m = 1,1+201= 2012 does not meet the requirements.

When m = 2, 2 2014 = 41007 = 41006× 4 =16 503× 4.

The last digit of 16 503 is 6 times 4, and the last digit is 4.

Obviously, m = 2 meets the requirements,

When m = 3, 3 2014 = 91007 = 91006× 9 = (81) 503× 9, which meets the requirements.

When m = 4, 4 2014 = (16)1007, and the last digit is 6, which does not meet the requirements.

When m = 5, the last digit of 5 20 14 is 5, which does not meet the requirements.

When m = 6, the last digit of 6 20 14 is 6, which does not meet the requirements.

When m = 7, 7 2014 = 491007 = 491006× 49 = (2401) 503× 49, and the last digit is 9, which meets the requirements.

When m = 8, 8 2014 = 641007 = 641006× 64 = 4096 503× 64, and the last digit is 4, which meets the requirements.

When m = 9, 9 20 1 4 = 811007, and the last one is1,which does not meet the requirements.

Therefore, the number of M that meets the requirements is 4, that is, M = 2, M = 3, M = 7 and M = 8.

Answer C.