Thus $n \le 12$. It's easy to construct such polygons with $5$-$12$ sides.

To complete the proof, we now show that $n$ cannot be $3$ or $4$. An interior square is a square with none of its vertices lie on the boundary (of the convex polygon). At a vertex of an interior square, there are either $2$ squares and $3$ triangles or $4$ squares. Thus there are at least $2$ squares at every vertex of an interior square. A boundary square is a square that has at least one vertex on the boundary. It is also easy to see that a boundary square must have one side on the boundary. We first show that the convex polygon must have a boundary square. Suppose on the contrary, all the squares are interior. Consider a line $L$ containing one side of the polygon. Let $v$ be a vertex belonging to a square so that $v$ is closest to $L$. At $v$ there are two squares. Since between $L$ and the line $M$ containing $v$ and parallel to $L$, there are no vertices belonging to a square, one side of each of the two squares containing $v$ must lie on $M$, i.e., the $2$ squares share a common side. Let $w$ be the last vertex on the 'left' side of $v$ on the line $M$ belonging to a square. By applying the same argument, there must be another vertex to the left of $w$ that belongs to a square, contradicting the choice of $w$. Therefore, there are boundary squares. If the boundary of the polygon formed is formed only by the edges of squares, we get a rectangle. This means the boundary is a cycle of squares. Removing these squares either leaves nothing or another rectangle. This means that no triangle is used, a contradiction. Thus on the boundary, there are edges belonging to triangles and squares. When an edge belonging to a triangle meets an edge belonging to a square, they meet at a vertex of the polygon and the angle at the vertex is $150^\circ$. There are at least two such vertices and therefore there are two angles which are $150^\circ$. This means $n$ cannot be $3$ or $4$.

1. The possible internal angles are $60^\circ$, $90^\circ$, $120^\circ$, $150^\circ$. Let their respective numbers be $a, b, c, d$. Then we have
$$\begin{aligned} a+b+c+d &= n \\ 60a+90b+120c+150d &= (n-2)180 \end{aligned}$$
Eliminating $n$, we obtain
$$4a+3b+2c+d=12.$$