2. Fill in each cell of an $n \times n$ grid paper with a number, such that each row and each column forms an arithmetic sequence. Such a filled grid paper is called an arithmetic cipher table. If knowing the numbers in certain cells of this arithmetic cipher table can decipher the entire table, then the set of these cells is called a key, and the number of cells in this set is called the length of the key.
(1) Find the smallest natural number $s$, such that any $s$ cells chosen from an $n \times n$ grid paper $(n \geqslant 4)$ form a key;
(2) Find the smallest natural number $t$, such that any $t$ cells chosen from the two diagonals of an $n \times n$ grid paper $(n \geqslant 4)$ form a key.
(Northwest University, Zhang Zhusheng, provided)
(1) When $s \leqslant 2 n-1$, take $s$ squares from the first row and the first column. Clearly, these $s$ squares do not form a key.

Below, we prove that when $s=2 n(n \geqslant 4)$, any $2 n$ squares taken from an $n \times n$ grid form a key. In $n$ columns, there are $2 n$ squares $(n \geqslant 4)$, so there must be two columns where each column has at least two squares (if $n-1$ columns have at most one square, then the total number of squares is at most $2 n-1$). Since it is an arithmetic password table, knowing two numbers $a_{i}, a_{j}(1 \leqslant i<j \leqslant n)$ in a column, the common difference $d=\frac{a_{j}-a_{i}}{j-i}$ can be calculated, and thus the column can be deciphered. Therefore, first, two columns can be deciphered, and then considering the $n$ rows, since two numbers in each row are known, similar to the column situation, each row can be deciphered. Thus, the entire arithmetic password table can be deciphered. Therefore, the smallest natural number $s = 2 n$.
(2) When $t \leqslant n$, take $t$ squares on the same diagonal, clearly, the arithmetic password table cannot be deciphered. When $t=n+1$, we prove below that the arithmetic password table can be deciphered. First, these $n+1$ squares are located in the $n$ columns of the grid, so there must be one column with at least two squares. Since each diagonal intersects each column at one square, there are only two squares in this column. From the proof in (1), this column can be deciphered. There are still $n-1$ squares $(n-1 \geqslant 3)$ not in this column. Since $n-1 \geqslant 3$, there are two possible scenarios: the first scenario is that there is another column with two squares, so this column can also be deciphered. With two columns deciphered, from the proof in (1), the entire arithmetic password table can be deciphered; the second scenario is that in the remaining $n-1$ columns, each column has only one square. Then consider the rows. Since each diagonal intersects each row at most at two points, these $n-1(n-1 \geqslant 3)$ squares are located in at least two rows, so at least two rows each have at least two known squares, one original and one deciphered from the column. Similar to the proof in (1), the entire password table can also be deciphered.