Problem 2309

Let $n \geq 5$ be a given integer. We are tasked with determining the greatest integer $k$ for which there exists a polygon with $n$ vertices (which can be either convex or non-convex, with a non-self-intersecting boundary) having $k$ internal right angles.

### Approach

To solve this problem, we must explore the possible configurations of polygons with given $n$ vertices and determine how many internal right angles can maximally exist. A right angle in a polygon is an internal angle equal to $90^\circ$.

### Steps

1. Basic Condition:
- For a polygon with $n$ vertices, the sum of interior angles is given by the formula:
$$\text{Sum of interior angles} = (n-2) \times 180^\circ.$$
- The presence of $k$ right angles implies $90k$ degrees are accounted out of the total sum:
$$90k + \text{Sum of remaining interior angles} = (n-2) \times 180^\circ.$$

2. Configuration of Remaining Angles:
- The remaining angles must also be non-right angles and satisfy the angle sum property. Often, this is managed by using angles greater than $90^\circ$ (making the rest of the polygon non-convex).

3. **Bounding $k$**:
- Let the remaining $n-k$ angles of the polygon be some angles that sum to the remaining degrees needed for the total angle sum.
- Since these remaining $n-k$ angles cannot be less than $0^\circ$ and angles greater than $180^\circ$ involve reflex angles which are invalid for simple (non-self-intersecting) polygons, a natural choice is using angles slightly greater than $90^\circ$, but less than or equal to $180^\circ$.

4. **Maximum $k$ Condition**:
- It can be shown through construction that it is feasible to achieve the maximum number by strategically using angles close to $135^\circ$ (non-right but feasible in self-boundary cases) for the remaining $n-k$ vertices.
- Thus, we estimate the maximum $k$ as:
$$90k + \approx 135(n-k) = (n-2) \times 180^\circ.$$
- Simplifying conditions for these remaining $n-k$ angles, and also ensuring positive counts of $135^\circ$ angles, gives the bounding function as:
$$k \approx \frac{2n + 1}{3}.$$

5. Final Integer Solution:
- Considering the largest integer that does not exceed the fraction, the closed form is found to be:
$$k = \left\lceil \frac{2n+1}{3} \right\rceil.$$

Thus, the greatest integer $k$ for which there exists such a polygon is:
$$\boxed{\left\lceil \frac{2n+1}{3} \right\rceil}$$