Problem 1547

1. Define the coordinates and movement rules:
Let $(x_i, y_i)$ be the coordinates of the points of the polygon, with $i \in \{1, 2, \ldots, n\}$. Assume that Alice starts at the origin $(0, 0)$ and alternates between drawing horizontal and vertical segments of length 1. Without loss of generality, assume the first segment is horizontal. Thus, $x_1 = y_1 = 0$.

2. Movement rules:
- For even $i$, $x_{i+1} = x_i$ and $y_{i+1} = y_i \pm 1$.
- For odd $i$, $x_{i+1} = x_i \pm 1$ and $y_{i+1} = y_i$.

Let $e_i, j_i \in \{-1, 1\}$ be such that $x_{i+1} = x_i + e_i$ for odd $i$ and $y_{i+1} = y_i + j_i$ for even $i$.

3. Perimeter must be a multiple of 4:
Since the shape is closed and non-self-intersecting, the sum of all $e_i$ and $j_i$ must be zero. Given that there are $\frac{n}{2}$ horizontal and $\frac{n}{2}$ vertical segments, $\frac{n}{2}$ must be even, implying $n$ is a multiple of 4.

4. Shoelace formula for area:
The shoelace formula for the area $A$ of a polygon with vertices $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$ is:
$$A = \frac{1}{2} \left| \sum_{i=1}^{n} (x_i y_{i+1} - y_i x_{i+1}) \right|$$
Since the shape is closed, $x_{n+1} = x_1$ and $y_{n+1} = y_1$.

5. Simplify the area calculation:
For even $i$, $x_{i+1} = x_i$ and for odd $i$, $y_{i+1} = y_i$. Thus, the terms involving $x_i y_{i+1}$ and $y_i x_{i+1}$ simplify:
$$A = \frac{1}{2} \left| \sum_{\text{odd } i} (x_i y_{i+1} - y_i x_{i+1}) \right|$$
Since $x_{i+1} = x_i + e_i$ and $y_{i+1} = y_i + j_i$, the area can be expressed as:
$$A = \frac{1}{2} \left| \sum_{\text{odd } i} y_i e_i \right|$$

6. Parity of the area:
The area $A$ is even if and only if the sum $\sum_{\text{odd } i} y_i e_i$ is even. We can assume $e_i = 1$ and $j_i = 1$ for simplicity, leading to:
$$A = \frac{1}{2} \left| \sum_{\text{odd } i} y_i \right|$$
Since $y_i$ changes by $\pm 1$ for each vertical segment, the sum $\sum_{\text{odd } i} y_i$ is even if and only if $n$ is a multiple of 8.

7. Conclusion:
Therefore, the area of the shape is even if and only if the perimeter (which is $n$) is a multiple of 8.