Find all integers $n \geq 3$ for which every convex equilateral $n$-gon of side length 1 contains an equilateral triangle of side length 1. We need to analyze the conditions such that any configuration of such a polygon will always have space to fit an equilateral triangle of unit side length.

### Analysis

1. Understanding the Geometry:
For an equilateral $n$-gon with side length 1, consider its symmetrical properties. The objective is to fit an equilateral triangle with side length 1 within this polygon, implying that our triangle shares a side or nearly a side with our polygon's perimeter.

2. Convex Equilateral Triangles:
- For $n = 3$, the polygon itself is an equilateral triangle, naturally containing itself as a unit triangle, satisfying the condition.
- For $n = 4$ (a square), it cannot necessarily contain an equilateral triangle of side 1 within it due to geometric restrictions, hence it cannot satisfy the condition.

3. **Odd-numbered $n$-gons**:
- When $n$ is odd, the symmetry and distribution of vertices allow the formation of such a triangle more easily due to alternative point distribution.
- By constructing examples for odd $n$, we observe it is always possible to fit an equilateral triangle within such polygons.

4. **Even-numbered $n$-gons**:
- Even configurations, such as squares, may lack the required space for such inclusions, especially due to vertex-edge distribution symmetry breaking and angle arrangements.

### Conclusion

Upon examining these geometric configurations, the restriction occurs primarily in even $n$-gons where the equilateral nature conflicts with containing a triangle of unit side as desired. However, for odd values of $n$, the structural properties allow an equilateral triangle of side length 1 to fit regardless of polygon orientation.

Thus, the set of $n$ that satisfy the problem's condition is:
$$\boxed{\text{all odd } n, \text{ } n \geq 3}$$