Neo would get lost in this Matrix

Configure the upper left $2014$ by $2014$ submatrix arbitrarily. Once we have a configuration, we can then fill the entries with indices $(i,2015)$ and $(2015,j)$ for $i,j = 1, 2, .... , 2014$ accordingly in order to satisfy the odd number of $1$ 's requirement for these rows and columns.

Once this is done, we can then fill the entry with index $(2015,2015)$ accordingly so that the $2015$ th row satisfies the "odd" requirement. But what happens to the $2015$ th column as a result? Now since we have filled $2015$ rows each with an odd number of $1$ 's there are an odd number of $1$ 's in the completed matrix. We have already filled $2014$ columns with an odd number of $1$ 's, so these $2014$ columns have an even number of $1$ 's. Thus to end up with an odd number of $1$ 's in total, the $2015$ th column must also have an odd number of $1$ 's, thus satisfying the given "odd" condition.

This "extension" of the $2014$ by $2014$ submatrix is unique, so any solution $2015$ by $2015$ matrix must correspond uniquely to one of these submatrices. We thus have a one-to-one correspondence, and since there are $2^{2014^{2}}$ submatrices, we have that $N = 2^{2014^{2}}.$ (This should be read as 2^(2014^2); the LaTeX gets quite small with an exponential "tower" like this.)

Therefore $(\log_{2} N) \pmod{1000} \equiv 2014^{2} \pmod{1000} \equiv 14^{2} \pmod{1000} = \boxed{196}.$

you may fill 2014 columns and rows in any way but last no is definite so as to make odd no of 1 in each row and column . so we get 2 power 2014 sqr whose mod(1000) is 196