The communication complexity of Hamming distance mod $4$
Consider the following task $f$: Given $x,y \in \{0,1\}^n$, Alice and Bob need to determine whether $d(x,y) \bmod{4} \in \{0,1\}$, where $d(x,y)$ is the Hamming distance between $x$ and $y$.

Let $M_f$ denote the matrix corresponding to this problem: $M_f(x,y) = 1$ if the answer is Yes, and $M_f(x,y) = -1$ if the answer is No. The discrepancy method shows that
$$
R_\epsilon(f) \geq \log \frac{1-2\epsilon}{\lambda_{\max}(M_f)/2^n}.
$$
In our case, you can calculate that $\lambda_{\max}(M_f) = 2^{\lceil n/2 \rceil}$ (I checked this for a few $n$ using Krawtchouk polynomials, but presumably there is a simple proof, since all the eigenvalues are nice). For even $n$, this implies that
$$
R_\epsilon(f) \geq \log \frac{1-2\epsilon}{2^{-n/2}} = \frac{n}{2} + \log(1-2\epsilon).
$$

(When $n$ is odd, the lower bound is slightly worse.)

Exactly the same bound is obtained if we replace the condition $d(x,y) \bmod{4} \in \{0,1\}$ with the condition $d(x,y) \bmod{4} \in \{0,3\}$.