3. To divide a polygon into a series of triangles using diagonals that do not intersect at interior points, it is necessary to take several points inside the shape as the vertices of the divided triangles when needed. These vertices are called "auxiliary vertices." The lines connecting auxiliary vertices and the lines connecting auxiliary vertices to vertices are all called diagonals. No two diagonals should intersect at an interior point. This method of dividing a polygon into a series of triangles is called the "triangulation" of the polygon.

Let $n$ be a positive integer not less than 5. To divide a regular $n$-sided polygon into a series of obtuse triangles, try to find the minimum possible value of the number of triangles $m$ formed.
3. The minimum possible value of $m$ is $n$.

If no auxiliary vertices are added, then all vertices of the divided triangles are on the $n$ vertices of the regular $n$-gon. Therefore, the circumcircle of the regular $n$-gon is the common circumcircle of all these triangles. Thus, they share a common circumcenter, which must lie inside or on the side of one of the divided triangles. Hence, that triangle is either an acute or a right triangle.

This indicates that if all the divided triangles are to be obtuse, auxiliary vertices must be added.

When the number of auxiliary vertices added is $k$, the sum of the interior angles of all the divided triangles is
$$
(n-2) \pi + 2k \pi = (n + 2k - 2) \pi,
$$

Thus, the number of divided triangles is $n + 2k - 2$. This shows that to divide the triangles into all obtuse triangles, the number of divided triangles $m \geqslant n$ (since $k \geqslant 1$). To achieve the minimum possible value of $m$, i.e., $m = n$, only one auxiliary vertex can be added.

Below, we show that adding just one auxiliary vertex is sufficient to achieve this.

First, use non-intersecting diagonals to divide the regular $n$-gon into $n-2$ triangles. Note that when $n$ is even, the main diagonals are not used. At this point, among the $n-2$ divided triangles, there are $n-3$ obtuse triangles and 1 acute triangle.

Then, take the incenter of the acute triangle as the auxiliary vertex and connect it to the three vertices of the acute triangle, thereby dividing the acute triangle into 3 obtuse triangles, resulting in a total of $n$ obtuse triangles.

When $n$ is even, let $n = 2l (l \geqslant 3)$. For the regular even-sided polygon $A_1 A_2 \cdots A_{2l}$, the specific operation is as follows: connect $A_1 A_3, A_1 A_4, \cdots, A_1 A_{2l-1}$, remove the main diagonal $A_1 A_{l+1}$, and instead connect $A_l A_{l+2}$, thus obtaining $n-2$ triangles. Since $l \geqslant 3$, $A_l A_{l+2}$ is not a main diagonal. At this point, the common circumcenter must lie within $\triangle A_1 A_l A_{l+2}$, so only this triangle is acute, and the other $n-3$ are obtuse. Then, take the incenter of $\triangle A_1 A_l A_{l+2}$ as the auxiliary vertex and connect it to $A_1, A_l, A_{l+2}$, thus obtaining $n$ obtuse triangles.