Example 1. Indicate one method for each of the following to divide a square with a side length of 1986 mm:
(1) Divide it into 1986 smaller squares with integer side lengths;
(2) Divide it into 1986 smaller squares with not all side lengths being integers.
(1) $\because 1986=2 \times 993$, divide the two sides $A B$ and $A D$ of the square into 993 equal parts (each part being $2 \text{ mm}$). As shown in Figure 5, using the discussion from Proof 1, we get: $2 \times 993-1=1985$ small squares with a side length of $2 \text{ mm}$, and one square $F C E E_{1}$ with a side length of $1984 \text{ mm}$. This results in 1986 integer-sided squares, and it is clear that this division method meets the requirements.
( 2) $\because 1986=662 \times 3$ is a $3k$ type number, referring to the discussion of $M_{2}$ in Proof 2: $1^{\circ}$ Perform one $B$-type division in the square, resulting in 5 smaller squares with a side length of $662 \text{ mm}$, and one square with a side length of $1324 \text{ mm}$, totaling 6 smaller squares.
$2^{\circ}$ Perform 662-2=660 $A$-type divisions. Since each $A$-type division reduces the side length of the resulting squares by half, after performing 660 $A$-type divisions in one of the 6 existing squares, the side length of the smallest square is $\frac{1}{2^{660}}$ of the original square's side length. Since $\frac{662}{2^{660}}$ and $\frac{1324}{2^{660}}$ are not integers, the result must include some non-integer-sided squares. At this point, one square is divided into $3 \times 660+1=1981$ smaller squares. Considering that one small square is used up during the $A$-type division, combining $1^{\circ}$ and $2^{\circ}$, we get: a square with a side length of $1986 \text{ mm}$ is divided into $6+1981-1=1986$ smaller squares, not all of which have integer sides.