PROOF: Fix X. Let i be an index such that Xi≠ 0.
For each vector R, let F(R) denote R with the ith bit flipped. Pair each vector R to its "partner" F(R). Since F(F(R)) = R, the partner relation is symmetric.
Now, for each pair of partners, (R, F(R)), exactly one of the two values {R.X, F(R).X} is 0 and the other is 1. Thus, half the vectors in Z2n have dot-product 1 with X, the remaining vectors have dot-product 0 with X.
QED
Note X.R = ∑i Xi Ri (mod 2).
Exercise: Generalize this to any finite field.