Problem 1383

1. Labeling the Sides:
John labels the sides of the regular $n$-gon with the numbers $1, 2, \ldots, n$ in some order. Let the labels be $a_1, a_2, \ldots, a_n$ in clockwise order.

2. Triangulation:
Mary divides the $n$-gon into $n-2$ triangles by drawing $n-3$ non-intersecting diagonals. Each diagonal is labeled with the number $1$.

3. Product Calculation:
For each triangle formed, the product of the numbers on its sides is written. Let the triangles be $\Delta_1, \Delta_2, \ldots, \Delta_{n-2}$, and let the sides of $\Delta_i$ be labeled with $a_{i1}, a_{i2}, a_{i3}$. The product for $\Delta_i$ is $a_{i1} \cdot a_{i2} \cdot a_{i3}$.

4. Sum Calculation:
Let $S$ be the sum of the products of the sides of all triangles:
$$S = \sum_{i=1}^{n-2} a_{i1} \cdot a_{i2} \cdot a_{i3}$$

5. Objective:
John wants to maximize $S$, while Mary wants to minimize $S$. Both will make the best possible choices.

6. Optimal Strategy for Mary:
Mary will try to minimize the sum $S$. To do this, she will aim to minimize the products of the sides of the triangles. Since the diagonals are labeled with $1$, the products will be minimized when the sides of the triangles are as small as possible.

7. Optimal Strategy for John:
John will try to maximize $S$. He will aim to place the largest numbers on the sides of the triangles as much as possible.

8. Analysis of the Products:
Consider the sum of the products of the sides of the triangles. Each side of the $n$-gon will appear in exactly two triangles. Therefore, the sum of the products can be written as:
$$S = \sum_{i=1}^{n} a_i \cdot (a_{i-1} + a_{i+1})$$
where $a_{i-1}$ and $a_{i+1}$ are the labels of the sides adjacent to $a_i$.

9. Maximizing the Sum:
To maximize $S$, John should place the largest numbers in positions where they will be multiplied by the largest possible sums of adjacent numbers.

10. Minimizing the Sum:
Mary will try to minimize the sum by choosing a triangulation that minimizes the products of the sides of the triangles. However, since each side appears in exactly two triangles, her influence is limited.

11. Conclusion:
Given that each side appears in exactly two triangles, the sum $S$ is determined by the sum of the products of the sides of the triangles. The optimal strategy for John is to place the numbers in such a way that the sum of the products is maximized.

The sum of the products of the sides of the triangles is:
$$S = \sum_{i=1}^{n} a_i \cdot (a_{i-1} + a_{i+1})$$
Since each number $1, 2, \ldots, n$ appears exactly twice, the sum $S$ is maximized when the numbers are placed in such a way that the products are maximized.

The final answer is $\boxed{n(n+1)}$