For all positive integers $n(n \geqslant 3)$, try to find those for which, after removing any one square from an $n \times n$ grid paper, the remaining grid paper can definitely be entirely cut into L-shaped pieces. Here, an L-shaped piece refers to a $2 \times 2$ grid paper piece with one small square removed, leaving the three remaining small squares.
When $3 \mid n$, by $3 \nmid\left(n^{2}-1\right)$, it is known that it does not meet the requirements.
When $3 \nmid n$, we discuss in cases.
(1) When $n=4$, we might as well remove a small square from the upper left $2 \times 2$ grid, and cut out the L-shape in Figure 1, reducing it to a $2 \times 2$ grid with one square removed, which can be entirely cut into L-shaped pieces. Therefore, the conclusion holds.
(2) When $n=5$, remove the 4th small square in the 1st row, then it is known that the two $2 \times 3$ grids A and B in Figure 2 must be cut using two L-shaped pieces each; and the remaining $3 \times 3$ grid cannot be entirely cut using L-shaped pieces. Therefore, it does not meet the requirements.
(3) When $n=7$, without loss of generality, we can assume that the removed square is within one of the four $2 \times 2$ grids I, II, III, IV in the upper left corner. By symmetry, we only need to consider grids I, II, and III, as shown in Figure 3.

Figures 4 to 6 respectively show the removed square in grids I, II, and III. The conclusion holds.
(4) When $n=8$, similar to the proof in case (1).
(5) When $n=10$, we might as well assume the removed square is in the upper left $8 \times 8$ grid, as shown in Figure 7, the conclusion holds.
(6) When $n=11$, we might as well assume the removed square is in the upper left $7 \times 7$ grid, as shown in Figure 8, the conclusion holds.
(7) When $n \geqslant 13$,

If $n=3 k+1$, similar to (5), from $n=3 k-1$ it can be extended to $3 k+1$;

If $n=3 k+2$, similar to (4), from $n=3 k-2$ it can be extended to $3 k+2$.
In summary, the required $n(n \geqslant 3)$ is $3+n, n \neq 5$.