Let S(i, j) be the switch at i-th row and j-th column. Let R(i) = 1 if i-th row has odd number of ‘off’s, otherwise R(i) = 0. Similarly define C(i) = 1 if j-th column has odd number of ‘off’s and C(i) = 0 if otherwise.

If N is even, any combination can be opened.
If N is odd, a necessary and sufficient condition for a combination to be opened is:
R(1) = R(2) = … = R(N) = C(1) = C(2) = … = C(N).
(So the example by pistons can not opened because R(1) = 1 and R(2) = 0.)

Proof:
If N is even, switch the status of all switches on i-th row and j-th column will change switch S(i, j) only and keep all other switches the same status. Repeating this process can turn on all switches.

If N is odd, changing any switch will change R(1), R(2), … R(N), C(1), … C(N). Since when the lock is opened, all R(i)s and C(i)s are zeros, a necessary condition is R(1) = R(2) = … = R(N) = C(1) = C(2) = … = C(N).

If this condition is satisfied, then this combination could be opened. It is obvious when N = 1. When N > 1, we can first turn on all switches from 2nd row to N-th row and 2nd column to N-th column (this is a N-1 by N-1 matrix, since N - 1 is even, this can be achieved). We still have R(1) = … = R(N) = C(1) = … = C(N) at this time. If all of them are zeros, then it must be the case that all switches are on. If they equal to 1, then all switches in the 1st row and 1st column are off. We only need change the switch at top-left corner to make all switches on.