Problem 928

Proof:
Sufficiency: It can be constructed, as shown in Figure $1-8$, a square can be dissected into 6 obtuse triangles, and any obtuse triangle can always be divided into two obtuse triangles.
Necessity: First, find the invariant relationship in any obtuse triangle dissection.
Classify the dissection points of the dissection method: dissection points on the boundary of the square or on a side of a dissection triangle but not at the vertex of the triangle are called first-class dissection points; dissection points not on the side of a triangle are called second-class dissection points.
In Figure 1-9, $F, G, H$ are first-class dissection points, and $E$ is a second-class dissection point.
For any obtuse triangle dissection of a square, it divides the square into $n$ obtuse triangles. Suppose this dissection has $r$ first-class dissection points and $s$ second-class dissection points. Since a first-class dissection point can provide at most one obtuse angle, and a second-class dissection point can provide at most three obtuse angles, we have $r+3 s \geqslant n$.
Calculating the total degree of the interior angles of the $n$ dissection triangles in two ways, we get
$$\begin{array}{c} n \times 180^{\circ}=r \times 180^{\circ}+s \times 360^{\circ}+4 \times 90^{\circ}, \\ n=r+2 s+2 . \end{array}$$

Therefore, $s \geqslant 2, n \geqslant 2 s+2 \geqslant 6$.