Assume $n$ is odd. Let $x_i^2+y_i^2=k$. From the obvious congruence $z^2 \equiv z \ (mod \ 2)$ so after summing the condition $b)$ we get $\Sigma x_i^2 + \Sigma y_i^2=k\cdot n \equiv \Sigma x_i + \Sigma y_i = 0 \equiv 0 \ (mod \ 2)$. So we get $2|k\cdot n$. By the assumption we have $2|k$. Looking at $b)$ we notice $x_i^2+y_i^2=k \equiv 0 \ (mod \ 2)$. Thus, $x_i \equiv y_i \ (mod \ 2)$ $(\star)$. Assume there exist $x_i$ and $x_j$ of different parity. We get $k=x_i^2+y_i^2 \equiv 0 \ (mod \ 4)$ but also $k=x_j^2+y_j^2 \equiv 2 \ (mod \ 4)$ hence all $x_i,y_i$ from $1$ to $n$ are of the same parity. If ( they are all odd ) { From $a)$ we get a sum of $n$ odd numbers to be even so $n$ must be even which is a contradiction } else { Look $v_2$. Let $\alpha_i, \beta_i$ be the power of $2$ fully dividing $x_i, y_i$ respectively. Also let $x_i=2^{\alpha_i} \cdot a_i$ and $y_i=2^{\beta_i} \cdot b_i$. Now WLOG $x_1$ has the lowest power of two fully dividing it. (keep in mind some might be equal). Obviously from the equation $x_1^2+y_1^2=k$ we get $2^{2 \alpha_1} | k$ so let $k=2^{2\alpha_1} \cdot r$. So we get $ a_1^2+2^{2\beta_1-2\alpha_1}=r$. Since $a_1$ is odd, we have another two cases. $1)$ $\beta_1=\alpha_1$ $2)$ $r$ is odd and thus $2^{2\alpha_1}$ fully divides $k$. And these two aren't hard.