Three. (50 points) Color some nodes (vertices of unit squares) on an $n \times n$ chessboard red, such that the boundary of any $k \times k (1 \leqslant k \leqslant n)$ sub-chessboard composed of unit squares has at least one red point. Let the minimum number of red points that satisfy this condition be $l(n)$. Try to find the value of $\lim _{n \rightarrow \infty} \frac{l(n)}{n^{2}}$.
For any red point $P$, assign a value to it as follows: if there are $m$ red points on the boundary of a unit square containing $P$, then $P$ gets a score of $\frac{1}{m}$ from that unit square. The value of point $P$ is the sum of the scores it gets from all unit squares containing it.

For red points on the boundary of the chessboard, the value of each point is at most 2. For a red point $P$ inside the chessboard, consider the $2 \times 2$ sub-chessboard centered at $P$. Its boundary must contain at least one other red point $Q$. For any unit square containing both $P$ and $Q$, $P$ gets a score of at most $\frac{1}{2}$ from it. Therefore, the value of point $P$ is at most $\frac{7}{2}$.

This shows that the value of any red point is at most $\frac{7}{2}$. Thus, the total value of $l(n)$ red points is at most $\frac{7}{2} l(n)$.

By the method of assignment, the total value of all red points on the chessboard should be $n^{2}$. Therefore,
$$
\frac{7}{2} l(n) \geqslant n^{2},
$$

which implies $\frac{l(n)}{n^{2}} \geqslant \frac{2}{7}$.
Consider a $\tilde{n} \times \tilde{n}$ chessboard, where $71(\tilde{n}+1)$, $n \leqslant \tilde{n} \leqslant n+6$.
Color the top-left $7 \times 7$ sub-chessboard as shown in Figure 4.
Extend this coloring method to the entire $\tilde{n} \times \tilde{n}$ chessboard, ensuring that the boundary of any $k \times k$ sub-chessboard contains at least one red point. In this coloring method, a total of $\frac{2}{7}(\tilde{n}+1)^{2}$ vertices are colored red.
Consider any $n \times n$ sub-chessboard of the $\tilde{n} \times \tilde{n}$ chessboard, we have
$$
l(n) \leqslant \frac{2}{7}(\tilde{n}+1)^{2} \leqslant \frac{2}{7}(n+7)^{2},
$$

which implies $\frac{l(n)}{n^{2}} \leqslant \frac{2}{7}\left(\frac{n+7}{n}\right)^{2}$.
Combining equations (1) and (2), we get
$$
\lim _{n \rightarrow+\infty} \frac{l(n)}{n^{2}}=\frac{2}{7} .
$$