Example 3 In the interior of square $ABCD$, there are 1999 points. Using the four vertices of the square and the 1999 interior points as vertices, it is divided into some triangles. How many triangles can be formed in total? How many cuts are needed in total?

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Solving from an overall perspective, first calculate the sum of the interior angles of all triangles.

In fact, the sum of the angles converging at a point inside a square is $360^{\circ}$, and the sum of the interior angles of the square is also $360^{\circ}$. Therefore, we have
$$
360^{\circ} \times 1999 + 360^{\circ} \text {. }
$$

Thus, the number of triangles is
$$
\frac{360^{\circ} \times 1999 + 360^{\circ}}{180^{\circ}} = 4000 \text {. }
$$

Since each triangle has three sides, and the original four sides of the square do not need to be cut, for the remaining sides, as they are shared by two triangles, cutting once results in two sides. Therefore, the total number of cuts is
$$
\frac{4000 \times 3 - 4}{2} = 5998 \text {. }
$$