Problem 1479

1. **For $n = 3$:**
- Consider a triangle. Any point inside or on the triangle will light up all three sides of the triangle. This is because the light from any point inside the triangle will reach all three sides without obstruction.
- Therefore, for $n = 3$, it is not possible to have a point inside the triangle such that each side has a point that is not lightened.

2. **For $n = 4$:**
- Consider a quadrilateral. Draw a diagonal that is fully contained inside the quadrilateral. This diagonal divides the quadrilateral into two triangles.
- Any light bulb placed inside one of these triangles will light up the sides of the quadrilateral that are part of that triangle. Hence, at least two sides of the quadrilateral will be fully lightened.
- Therefore, for $n = 4$, it is not possible to have a point inside the quadrilateral such that each side has a point that is not lightened.

3. **For $n = 5$:**
- Consider a pentagon. Draw a diagonal that is fully contained inside the pentagon. This diagonal divides the pentagon into a triangle and a quadrilateral.
- If the light bulb is placed inside the triangle, it will light up the sides of the pentagon that are part of the triangle. If it is placed inside the quadrilateral, it will light up at least two sides of the quadrilateral, and hence at least one side of the pentagon.
- Therefore, for $n = 5$, it is not possible to have a point inside the pentagon such that each side has a point that is not lightened.

4. **For $n = 6$:**
- Consider a hexagon. It is possible to place a light bulb at a point $P$ inside the hexagon such that each side of the hexagon has a point that is not lightened. This can be achieved by placing the light bulb at a point that is not equidistant from all sides, creating shadows on each side.
- Therefore, $n = 6$ is the smallest integer for which there exists an $n$-gon and a point within it such that, if a light bulb is placed at that point, on each side of the polygon there will be a point that is not lightened.

5. Second Part:
- For $n = 6$, we need to show that there always exist two points within the hexagon such that the bulbs placed at these points will lighten up the whole perimeter of the hexagon.
- This can be done by recalling the Art Gallery Theorem, which states that for any simple polygon with $n$ vertices, $\left\lfloor \frac{n}{3} \right\rfloor$ guards (or light bulbs) are sufficient to cover the entire polygon.
- For $n = 6$, $\left\lfloor \frac{6}{3} \right\rfloor = 2$. Therefore, two light bulbs are sufficient to light up the entire perimeter of the hexagon.

The final answer is $\boxed{ n = 6 }$.