Let $n = 4k$. For each $i = 1, 2, \dots, n$, denote $\alpha_i = i \cdot \frac{360^\circ}{n}$ and $x_i = \sin^2 \alpha_i$. We pair the terms as $(x_1, x_{k+1}), (x_2, x_{k+2}), \dots, (x_k, x_{2k})$ and $(x_{2k+1}, x_{3k+1}), (x_{2k+2}, x_{3k+2}), \dots, (x_{3k}, x_{4k})$. Since
$$\alpha_{i+k} = (i + k) \cdot \frac{360^\circ}{n} = i \cdot \frac{360^\circ}{n} + k \cdot \frac{360^\circ}{n} = \alpha_i + 90^\circ,$$
we have
$$\sin^2 \alpha_{i+k} = \sin^2 (\alpha_i + 90^\circ) = \sin^2 (90^\circ - (-\alpha_i)) = \cos^2 (-\alpha_i) = \cos^2 \alpha_i.$$
Therefore $x_i + x_{i+k} = \sin^2 \alpha_i + \cos^2 \alpha_i = 1$. Thus, the sum of the numbers of each pair is $1$. Since we have $\frac{n}{2}$ pairs, the sum of all numbers is $\frac{n}{2}$.

Let $n = 4k$. Note that
$$\sin^2 \left( i \cdot \frac{360^\circ}{n} \right) = \frac{1 - \cos \left( 2i \cdot \frac{360^\circ}{n} \right)}{2} = \frac{1}{2} - \frac{1}{2} \cos \left( 2i \cdot \frac{360^\circ}{n} \right).$$
For each $i = 1, 2, \dots, n$, denote $\alpha_i = i \cdot \frac{360^\circ}{n}$ and $y_i = \cos 2\alpha_i$; then the equation to be proven takes the form $\frac{1}{2}n - \frac{1}{2}(y_1 + y_2 + \dots + y_n) = \frac{1}{2}n$, which is equivalent to $y_1 + y_2 + \dots + y_n = 0$. We are going to prove this equality in the rest. We pair the terms as $(y_1, y_{k+1}), (y_2, y_{k+2}), \dots, (y_k, y_{2k})$ and $(y_{2k+1}, y_{3k+1}), (y_{2k+2}, y_{3k+2}), \dots, (y_{3k}, y_{4k})$ and prove the angle equality $\alpha_{i+k} = \alpha_i + 90^\circ$ as in Solution 1. Next, we see that
$$\cos 2(\alpha_i + 90^\circ) = \cos (2\alpha_i + 180^\circ) = -\cos 2\alpha_i,$$
from which it follows that $y_i + y_{i+k} = \cos 2\alpha_i - \cos 2\alpha_i = 0$. Since the sum of the numbers of each pair is $0$, the sum of all numbers is also $0$.