4. In an equilateral $\triangle ABC$, a ray is drawn from vertex $A$ into the interior of the triangle, such that it reaches a side. When the ray reaches a side, it bounces off according to the reflection rule, i.e., if it arrives at an oriented angle $\alpha$, it leaves at an oriented angle $180^{\circ}-\alpha$. After $n$ reflections, the ray returns to point $A$, and does not pass through the other two vertices. Find all possible values of $n$.
4. All positive integers $n \equiv 1,5(\bmod 6)$ except $5$ and $17$ satisfy the condition.

Consider an equilateral triangle $\triangle A A_{1} A_{2}$ with side length $m$, and perform a unit length triangular subdivision, as shown in Figure 5.

Assign a pair of coordinates $(a, b)$ to each vertex obtained from the subdivision, where $a$ and $b$ are the number of edges traversed in the directions of $A A_{1}$ and $A A_{2}$, respectively, such as
$$
A=(0,0), A_{1}=(m, 0), A_{2}=(0, m) .
$$

Denote the unit triangle with vertex $A$ as $\triangle A B C$, where $B=(1,0)$ and $C=(0,1)$.

Each unit triangle can be obtained from $\triangle A B C$ through a series of edge reflections. For example, the vertex $(1,1)$ is the reflection of point $A$ about $B C$, and the vertex $(0,2)$ is the reflection of point $B$ about the edge formed by points $C$ and $(1,1)$.

Notice that in the reflection of vertex $(a, b)$ about an edge, $a-b(\bmod 3)$ remains unchanged, and vertices $(a, b)$ and $\left(a^{\prime}, b^{\prime}\right)$ can be mutually reached through a series of such reflections if and only if
$$
a-b \equiv a^{\prime}-b^{\prime}(\bmod 3).
$$
Thus, the points $(a, b)$ obtained from $A$ through reflections satisfy $a \equiv b(\bmod 3)$. Denote the set of such points as $V$.

Let $U$ be the set of vertices that are not on any line segment from $A$ to any other vertex.
Then $(a, b) \in U$ if and only if $(a, b)=1$.
Otherwise, if $d=(a, b)>1$, then the vertex $\left(\frac{a}{d}, \frac{b}{d}\right)$ is also on the line segment connecting point $(a, b)$ and $A$, which is a contradiction.

Notice that the rays from $A$ that eventually return to $A$ are the line segments from $A$ to points in $V \cap U$. In Figure 5, the number of lines intersected by the ray to such a point $(a, b)$ is
$$
(a-1)+(b-1)+(a+b-1)=2(a+b)-3
$$

lines, where $a-1$ lines are parallel to $A B$, $b-1$ lines are parallel to $A C$, and $a+b-1$ lines are parallel to $B C$.
Thus, the ray reflects $2(a+b)-3$ times in $\triangle A B C$ before returning to $A$. Therefore, the problem reduces to finding all $n=2(a+b)-3$ that satisfy
$$
a \equiv b(\bmod 3) \text { and }(a, b)=1.
$$

It is easy to see that $a+b$ is not a multiple of 3 and $n$ is odd.
Thus, $n \equiv \pm 1(\bmod 6)$.
By taking $(1,3 k+1)$, we get $n=6 k+1(k \geqslant 0)$;
By taking $(a, b)=(3 k-1,3 k+5)$ (for even $k$) or $(3 k-4,3 k+8)$ (for odd $k$), we get all $n=12 k+5(k \geqslant 2)$;
By taking $(a, b)=(3 k-1,3 k+2)$, we get all $n=12 k-1(k \geqslant 1)$.
For $n=5$ and $n=17$, we only need to consider the equations
$2(a+b)-3=5$ and $2(a+b)-3=17$.
Clearly, there are no solutions where $a \equiv b(\bmod 3)$ and $(a, b)=1$.