Problem 1207

$$\begin{array}{lc} \text { II. Let } & \alpha_{i}=2 i-1, \beta_{i}=201-\alpha_{i}(i=1,2, \cdots, 100) \\ \text { and } & E_{i}=\left\{\alpha_{i}, \beta_{i}\right\}, \end{array}$$

Then when $i \neq j$, $E_{i} \cap E_{j}=\varnothing$, and $\bigcup_{i=1}^{100} E_{i}=E$.
From i) we know that for any $1 \leqslant i \leqslant 100$, it cannot be that $E_{i} \subset G$.
Furthermore, since the number of elements in $G$ is exactly equal to the number of sets $E_{i}$, which is 100, $G$ must necessarily contain exactly one element from each $E_{i}$.
Now suppose $G$ contains $k$ odd numbers, and for
$$1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant 100,$$

we have
$$a_{i_{t}}=\alpha_{i_{t}}(1 \leqslant t \leqslant k),$$

Thus, for $\quad j \neq i_{i}(1 \leqslant t \leqslant k)$, it must be that $a_{j}=\beta_{j}$
From ii)
$$\sum_{t=1}^{k} \alpha_{i t}+\sum_{j \neq i_{t}} \beta_{j}=10080$$

On the other hand
$$\sum_{j=1}^{100} \beta_{j}=2 \sum_{j=1}^{100} j=10100$$
Subtracting (1) from (2)
$$\begin{array}{c} \sum_{t=1}^{k}\left(\beta_{i t}-\alpha_{i_{i}}\right)=20 \\ k \cdot 201-2 \sum_{t=1}^{k} \alpha_{i_{i}}=20 \\ \sum_{t=1}^{k} \alpha_{i t}=\frac{1}{2}(k \cdot 201-20) \end{array}$$

From (4), we first deduce that $k$ must be a positive even number. Let $k=2 k^{\prime}$, then $\sum_{k=1}^{2 k^{\prime}} \alpha_{i t}=k^{\prime} \cdot 201-10$. Noting that the left side of this equation is even, $k^{\prime}$ must be a positive even number. Let $k^{\prime}=2 \cdot k^{\prime \prime}$, then $k=4 k^{\prime \prime}$, which proves that the number of odd numbers in $G$ must be a multiple of 4.
Now we calculate the sum of the squares of the numbers in $G$. The following calculation uses (3):
$$\begin{aligned} \sum_{i=1}^{100} \alpha_{i}^{2} & =\sum_{t=1}^{k} \alpha_{i_{t}}^{2}+\sum_{j \neq i_{t}}^{k} \beta_{j}^{2} \\ & =\sum_{j=1}^{100} \beta_{j}^{2}-\sum_{t=1}^{k} \beta_{i_{i}}^{2}+\sum_{t=1}^{k} \alpha_{i_{t}}^{2} \\ & =\sum_{j=1}^{100}(2 j)^{2}-\sum_{t=1}^{k}\left(\beta_{i_{t}}+\alpha_{i_{t}}\right)\left(\beta_{i_{t}}-\alpha_{i_{t}}\right) \\ & =4 \sum_{j=1}^{100} j^{2}-201 \times 20 \\ & =4 \times \frac{100(100+1)(200+1)}{6}-4020 \\ & =1353400-4020=1349380 . \end{aligned}$$