2. Let $n \geqslant 2$ be an integer. Consider an $n \times n$ chessboard composed of $n^{2}$ unit squares. If each row and each column contains exactly one “rook”, then a placement of $n$ rooks is called “peaceful”. Find the largest positive integer $k$, such that for any peaceful placement of $n$ rooks, there exists a $k \times k$ square, whose $k^{2}$ unit squares all do not contain a rook.
2. $[\sqrt{n-1}]$.

Let $l$ be a positive integer. Next, we prove two conclusions, thereby illustrating that the maximum value sought is $k_{\text {max }}=[\sqrt{n-1}]$.
(1) If $n>l^{2}$, then for any peaceful placement of $n$ rooks, there exists an empty $l \times l$ square.
(2) If $n \leqslant l^{2}$, then there exists a peaceful placement of $n$ rooks such that each $l \times l$ square contains at least one rook.
Proof (1) When $n>l^{2}$, for any peaceful placement of $n$ rooks, there exists a row $R$ such that the first square in this row contains a rook. Take $l$ consecutive rows (including row $R$), and their union $U$ contains exactly $l$ rooks. At this point, consider the 2nd to the $l^{2}+1(\leqslant n)$ columns in set $U$, which form an $l \times l^{2}$ rectangle that can be divided into $l$ $l \times l$ squares. However, there are at most $l-1$ rooks in these squares (since set $U$ already has one rook in the 1st column). Therefore, at least one of these $l$ $l \times l$ squares is empty.
(2) For the case $n=l^{2}$, first construct a peaceful placement without any empty $l \times l$ square. Label all rows from bottom to top as $0,1, \cdots, l^{2}-1$, and label all columns from left to right as $0,1, \cdots, l^{2}-1$, and denote the unit square at the $r$-th row and $c$-th column as $(r, c)$.
Place a rook in each unit square of the form $(i l+j, j l+i)(i, j=0,1, \cdots, l-1)$.
Figure 1 shows the placement when $l=3$.
Figure 1
Since each number from 0 to $l^{2}-1$ has a unique representation of the form $i l+j(0 \leqslant i, j \leqslant l-1)$, each row and each column contains exactly one rook.
At this point, for any $l \times l$ square $A$, let the label of the lowest row of square $A$ be $p l+q(0 \leqslant p, q \leqslant l-1$, because $\left.p l+q \leqslant l^{2}-l\right)$. Then, the $l$ rows of square $A$ contain $l$ rooks, and the labels of the columns where these $l$ rooks are located are
$$
\begin{array}{l}
q l+p,(q+1) l+p, \cdots,(l-1) l+p, p+1, \\
l+(p+1), \cdots,(q-1) l+p+1,
\end{array}
$$

Arrange these numbers in ascending order:
$$
\begin{array}{l}
p+1, l+(p+1), \cdots,(q-1) l+p+1, \\
q l+p,(q+1) l+p, \cdots,(l-1) l+p .
\end{array}
$$

The difference between any two adjacent numbers in this sequence does not exceed $l$, the first number does not exceed $l-1$ (if $p=l-1$, then $q=0$, so the first number will be $q l+p=l-1$), and the last number is at least $(l-1) l$. Therefore, at least one of the labels of the $l$ columns of square $A$ appears in this sequence, indicating that square $A$ contains at least one rook. This completes the construction for $n=l^{2}$.

Next, construct a peaceful placement without any empty $l \times l$ square for the case $n<l^{2}$.

For this, consider the construction for the $l^{2} \times l^{2}$ case as described above, and delete the rightmost $l^{2}-n$ columns and the bottom $l^{2}-n$ rows, resulting in an $n \times n$ square, and there are no empty $l \times l$ squares in it. However, some rows and columns may be empty, and the number of empty rows is equal to the number of empty columns. Therefore, these rows and columns can be paired, and a rook can be placed at the intersection of the corresponding empty row and column. Thus, a peaceful placement of $n$ rooks on an $n \times n$ board is obtained, and there are no empty $l \times l$ squares.
In summary, the maximum value sought is $k_{\max }=[\sqrt{n-1}]$.