$7 \cdot 46$ A cube with edge length $n$ is divided into $n^{3}$ (edge length 1) unit cubes. Select some of the small cubes and draw 3 lines parallel to the edges through the center of each selected small cube. How many small cubes must be selected at a minimum to ensure that the lines through their centers cover all the small cubes?
(1) For $n=2,3,4$, answer the above question.
(2) Find the answer for $n=10$.
(3) Solve the general problem. If an exact answer cannot be found, provide an estimate inequality for the number of small cubes and prove it.
【Solution】The right figure shows the bottom faces of four types of cubes, where the numbers filled in indicate the layer number of the selected small cube above the square. This means that for cubes with edge lengths of $2,3,4,5$, it is sufficient to select $2,5,8,13$ small cubes, respectively. Moreover, it can be proven that these numbers cannot be reduced. Therefore, we can conjecture that for a general $n$, the minimum number of small cubes to be selected is
$$
A_{n}=\left\{\begin{array}{ll}
\frac{n^{2}}{2}, & \text { when } n \text { is even, } \\
\frac{n^{2}+1}{2}, & \text { when } n \text { is odd. }
\end{array}\right.
$$

To prove this conclusion, we first provide the following lemma:
Lemma In an $n \times n$ grid, each cell is filled with a non-negative integer. For any zero in the grid, if the sum of the other $2n-2$ numbers in its row and column is not less than $n$, then the sum of all numbers in the grid is less than $\frac{n^{2}}{2}$.

Proof of the lemma: Sum the numbers in each row and each column of the grid, and let the minimum of these $2n$ sums be $m$. If $m \geqslant \frac{n}{2}$, then the sum of all numbers in the grid $S \geqslant \frac{n^{2}}{2}$. Assume $m<\frac{n}{2}$.

In this case, without loss of generality, assume the sum of the numbers in the first row is $m$, and the first $q$ elements in this row are non-zero, while the last $n-q$ elements are zero. Clearly, $q \leqslant m<\frac{n}{2}$. According to the given condition, the sum of the numbers in each of the last $n-q$ columns is not less than $n-m$, and the sum of the numbers in each of the first $q$ columns is not less than $m$, hence we have
$$
\begin{aligned}
S & \geqslant q m+(n-q)(n-m)=\frac{n^{2}}{2}+\frac{1}{2}(n-2 q)(n-2 m) \\
& \geqslant \frac{n^{2}}{2} .
\end{aligned}
$$

Now we prove the conclusion (1). Suppose a set of cubes that meet the requirements has already been selected. Now, fill each cell of the $n \times n$ grid on the bottom face of the large cube with a non-negative integer, representing the number of selected small cubes in the $n$ small cubes above this cell. At this point, if there is a cell with a number 0, it means that none of the $n$ cubes above this cell have been selected. Therefore, each of these cubes must be eliminated by a selected small cube in the same layer that is in the same row or column. This implies that in the $n \times n$ number grid on the bottom face, the sum of the other $2n-2$ numbers in the row and column of any cell with a 0 is not less than $n$. Thus, by the lemma, the sum of all numbers in the grid is not less than $\frac{n^{2}}{2}$, which means the number of selected small cubes is not less than $A_{n}$:

For a general $n$, it is also possible to construct an example of selecting $A_{n}$ small cubes that meet the requirements, similar to the examples given at the beginning.

In summary, for each $n \geqslant 2$, the minimum number of small cubes to be selected to meet the requirements is $A_{n}$, given by (1). Specifically, we have $A_{2}=2, A_{3}=5, A_{4}=8, A_{10}=50$.