Problem 1489

1. **Assume $n$ is odd.** Let $x_i^2 + y_i^2 = k$ for all $i$. From the given conditions, we have:
$$\sum_{i=1}^n x_i = \sum_{i=1}^n y_i = 0$$
and
$$x_1^2 + y_1^2 = x_2^2 + y_2^2 = \cdots = x_n^2 + y_n^2 = k.$$

2. Consider the congruence modulo 2. Note that $z^2 \equiv z \pmod{2}$ for any integer $z$. Therefore, summing the squares, we get:
$$\sum_{i=1}^n x_i^2 + \sum_{i=1}^n y_i^2 = k \cdot n \equiv \sum_{i=1}^n x_i + \sum_{i=1}^n y_i = 0 \equiv 0 \pmod{2}.$$
This implies:
$$k \cdot n \equiv 0 \pmod{2}.$$

3. **Since $n$ is odd, $k$ must be even.** This is because the product of an odd number and an even number is even. Therefore, $k \equiv 0 \pmod{2}$.

4. **Analyze the parity of $x_i$ and $y_i$.** Since $x_i^2 + y_i^2 = k \equiv 0 \pmod{2}$, both $x_i$ and $y_i$ must have the same parity (either both are even or both are odd). Assume there exist $x_i$ and $x_j$ of different parity. Then:
$$k = x_i^2 + y_i^2 \equiv 0 \pmod{4}$$
but also
$$k = x_j^2 + y_j^2 \equiv 2 \pmod{4},$$
which is a contradiction. Hence, all $x_i$ and $y_i$ must have the same parity.

5. **Consider the case where all $x_i$ and $y_i$ are odd.** From condition (a), the sum of $n$ odd numbers must be even, which implies $n$ must be even. This contradicts our assumption that $n$ is odd.

6. **Consider the case where all $x_i$ and $y_i$ are even.** Let $x_i = 2^{\alpha_i} \cdot a_i$ and $y_i = 2^{\beta_i} \cdot b_i$, where $a_i$ and $b_i$ are odd. Without loss of generality, assume $x_1$ has the lowest power of 2 fully dividing it. From $x_1^2 + y_1^2 = k$, we get:
$$2^{2\alpha_1} \mid k.$$
Let $k = 2^{2\alpha_1} \cdot r$. Then:
$$a_1^2 + 2^{2\beta_1 - 2\alpha_1} b_1^2 = r.$$
Since $a_1$ is odd, we have two cases:
1. $\beta_1 = \alpha_1$.
2. $r$ is odd, and thus $2^{2\alpha_1}$ fully divides $k$.

Both cases lead to a contradiction if $n$ is odd.

Therefore, $n$ must be even.

$\blacksquare$