Expanding and clarifying Ady's methodology:

Let t=M-N 

Then M=(2013-3t)/(t-1) = 2010/(t-1) - 3

M is positive if t is any factor of 2010 less than 670 (i.e. 2010/3).

2010 = 2*3*5*67, so its factors under 670, all of which give a positive M, are 1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402;

However, in order for N to also be positive, t must be less than M.  sqrt(2010) is about 45, so there are only 8 values of (t-1) that give both a positive M and N, namely 1, 2, 3, 5, 6, 10, 15, 30.  This leads to Ady's correct answer.

     for t= 2, 3,4,6,7,11, 16 and 31;  

     producing 8 possible couples for  (M,N):

      (2007,2005),  (1002,999),  (667,663), (399,393), 

      (332,325,)  (198,187), (131,115), (64,33)

Actually, Ady's last pair contained a typo, which I have corrected in my answer.