11.57 In an $n \times n$ grid, each cell is filled with a number such that the numbers in each row and each column form an arithmetic sequence. Such a filled grid is called an arithmetic cipher table. If the numbers in certain cells of this arithmetic cipher table are known, the cipher table can be deciphered, and these cells are called a key. 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 in an $n \times n (n \geqslant 4)$ grid form a key;
(2) Find the smallest natural number $t$, such that any $t$ cells chosen from the two diagonals of an $n \times n (n \geqslant 4)$ grid form a key.
[Solution] (1) First, when it is known that the numbers in the $2n-1$ squares of the 1st row and the 1st column of the grid paper are all 0, the arithmetic password table cannot be uniquely determined. At this time, the number table where all $n \times n$ squares are 0 and the following number table are both arithmetic:
\begin{tabular}{cccccl}
0, & 0, & 0, & 0, & $\cdots$, & 0 \\
0, & 1, & 2, & 3, & $\cdots$, & $n-1$ \\
0, & 2, & 4, & 6, & $\cdots$, & $2(n-1)$ \\
0, & 3, & 6, & 9, & $\cdots$, & $3(n-1)$ \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & & $\vdots$ \\
0, & $n-1,2(n-1), 3(n-1)$, & $\cdots$, & $(n-1)^{2}$.
\end{tabular}

This indicates that the smallest natural number $s \geqslant 2n$.
On the other hand, for any fixed $2n$ squares, if the numbers in these $2n$ squares are known, then by the pigeonhole principle, there must be at least two fixed squares in one row, thus the $n$ numbers in this row can be determined. Since there are at most $n$ fixed squares in this row, there are at least $n$ fixed squares in the remaining $n-1$ rows, so there must be at least two fixed squares in one row. Thus, the numbers in this row are also determined. Since two rows of numbers have been determined, it is known that the password table is uniquely determined, i.e., any $2n$ squares are a key. Therefore, the smallest natural number $s$ is $2n$.
(2) When a diagonal of $n$ squares is fixed, and the numbers in these $n$ squares are all 0, the number table where all squares are 0 and the following number table both meet the requirements of the problem:
\begin{tabular}{ccccc}
0, & 1, & 2, & $\cdots$, & $n-1$ \\
-1, & 0, & 1, & $\cdots$, & $n-2$ \\
-2, & -1, & 0, & $\cdots$, & $n-3$ \\
$\vdots$ & $\vdots$ & $\vdots$ & & $\vdots$ \\
$-(n-1)$, & $-(n-2)$, & $-(n-3)$, & $\cdots$, & 0.
\end{tabular}