Problem 1698

1. Divide the sequences: Given the sequences $\{a_i\}_{1 \leq i \leq n}$ and $\{b_i\}_{1 \leq i \leq n}$, we can divide them into two subsets each with equal sums. Specifically, we divide $\{a_i\}$ into $\{x_i\}_{1 \leq i \leq k}$ and $\{y_i\}_{1 \leq i \leq n-k}$ such that:
$$\sum_{i=1}^{k} x_i = \sum_{i=1}^{n-k} y_i$$
Similarly, we divide $\{b_i\}$ into $\{z_i\}_{1 \leq i \leq l}$ and $\{w_i\}_{1 \leq i \leq n-l}$ such that:
$$\sum_{i=1}^{l} z_i = \sum_{i=1}^{n-l} w_i$$

2. Order the subsets: Arrange the elements in each subset in non-increasing or non-decreasing order:
$$x_1 \geq x_2 \geq \dots \geq x_k, \quad y_1 \leq y_2 \leq \dots \leq y_{n-k}$$
$$z_1 \geq z_2 \geq \dots \geq z_l, \quad w_1 \leq w_2 \leq \dots \leq w_{n-l}$$

3. Define cumulative sums: Define the cumulative sums for each subset:
$$X_j = \sum_{i=1}^{j} x_i, \quad Y_j = \sum_{i=1}^{j} y_i$$
$$Z_j = \sum_{i=1}^{j} z_i, \quad W_j = \sum_{i=1}^{j} w_i$$

4. Construct the polygon: Construct a $2n$-polygon $P_1P_2 \dots P_{2n}$ with vertices defined as follows:
$$P_{2i-1} = (X_i, W_{i-1}), \quad P_{2i} = (X_i, W_i) \quad \text{for } 1 \leq i \leq k$$
$$P_{2i-1} = (Y_{n-i}, W_{i-1}), \quad P_{2i} = (Y_{n-i}, W_i) \quad \text{for } k+1 \leq i \leq n-l$$
$$P_{2i-1} = (Y_{n-i}, Z_{n-i+1}), \quad P_{2i} = (Y_{n-i}, Z_{n-i}) \quad \text{for } n-l+1 \leq i \leq n$$

5. Prove the polygon is simple: To show that the polygon $P_1P_2 \dots P_{2n}$ is simple (i.e., it does not intersect itself), we need to verify the following conditions:
$$\frac{W_1}{X_1} \leq \frac{W_2}{X_2} \leq \dots \leq \frac{W_k}{X_k} \leq \frac{Z_l}{Y_l} \leq \frac{Z_{l-1}}{Y_{l-1}} \leq \dots \leq \frac{Z_1}{Y_1}$$
$$\frac{W_{k-1}}{X_{k-1}} < \frac{Z_{l-1}}{Y_{l-1}}$$

6. Verify condition (1): The first condition is straightforward to verify given the ordering of the subsets and the properties of the cumulative sums.

7. Verify condition (2): Suppose the second condition is false. By calculation, this would imply that all $a_i$'s are equal and all $b_i$'s are equal, which contradicts the given condition that $a_1, a_2, \dots, a_n$ are not all equal.

Thus, the conditions are satisfied, and the constructed polygon is simple.

$\blacksquare$