Example 3 A $7 \times 7$ grid is completely covered without overlap by 16 $1 \times 3$ rectangles (which can be rotated $90^{\circ}$). Question: In what positions will the small squares that are not covered be?
Solve: Divide a $7 \times 7$ grid into 49 small squares, and label them with coordinates similar to Example 1.

Assume the square that cannot be covered is $(a, b)$. Label the square $(i, j)$ with $\varepsilon^{i+j}$, where $\varepsilon=\mathrm{e}^{\frac{2 \pi}{3}}$ is a cube root of unity.

Clearly, the sum of the numbers in the squares of a $1 \times 3$ or $3 \times 1$ rectangle is 0, so the number in the square $(a, b)$ is the sum of the numbers in the grid, i.e.,
$$
\varepsilon^{a+b}=\left(\sum_{i=1}^{7} \varepsilon^{i}\right) \sum_{j=1}^{7} \varepsilon^{j}=\varepsilon^{1} \cdot \varepsilon^{1} \text {. }
$$

Thus, $a+b \equiv 2(\bmod 3)$.
Similarly, if the square $(i, j)$ is labeled with $\varepsilon^{i+2 j}$, then the sum of the numbers in the squares of a $1 \times 3$ or $3 \times 1$ rectangle is 0.
$$
\begin{array}{l}
\text { Hence } \varepsilon^{a+2 b}=\left(\sum_{i=1}^{7} \varepsilon^{i}\right) \sum_{j=1}^{7} \varepsilon^{2 j} \\
=\varepsilon \cdot \frac{\varepsilon^{7}-1}{\varepsilon-1} \cdot \varepsilon^{2} \cdot \frac{\varepsilon^{14}-1}{\varepsilon^{2}-1}=\varepsilon^{3}=1 .
\end{array}
$$

Thus, $a+2 b \equiv 0(\bmod 3)$.
In summary, $a \equiv b \equiv 1(\bmod 3)$.
Therefore, $(a, b)$
$$
\begin{aligned}
= & (1,1),(1,4),(1,7),(4,1),(4,4), \\
& (4,7),(7,1),(7,4),(7,7) .
\end{aligned}
$$

Thus, the square can only be at the four corners, the midpoints of the four edges, or the center of the entire grid.
Here, it is necessary to provide examples for each case to demonstrate feasibility.
It is easy to find that examples for the four corners and the midpoints of the four edges are very straightforward, while for the center of the entire grid, a rotational construction method (as shown in Figure 1) is needed.

Of course, if the proposition is changed to other irregular shapes or the covered area is not a square grid, the problem can be handled by appropriately selecting the unit roots.