Let $m$ be the sum, over all pairs of days, of the number of people who give the a different answer on the two days. Then $m = \frac{nk (k-1)}{4}$. For any person who gives an equal number of yeses and nos, the number of pairs of days on which he gives a different answer is $\frac{k^{2}}{4}$. So the maximum number of such people is: $\lfloor \frac{\frac{nk (k-1)}{4}}{\frac{k^{2}}{4}}\rfloor$ = $\lfloor n-\frac{n}{k}\rfloor$.

Let $ v_i$ be the vector that $ v_i=(a_{i1},\ldots,a_{in})$ where $ a_{ij}=1$ if the j-th person's answer on the i-th day is yes and $ a_{ij}=-1$, otherwise. The giving condition implies that $ v_i \cdot v_j=0$ and clearly $ v_i^2=n$. Let the i-th person's answer (1 or -1) adds up to $ x_i$. If there is more than $ n- \frac{n}{k}$ people whose yes's equal to no's then $ k^2 \cdot \frac{n}{k} < (x_1^2+\ldots+x_n^2)=|v_1+\ldots+v_k|^2=v_1^2+\ldots+v_k^2=kn$. Contradiction.