4. In an $n \times n$ grid, some cells are colored black, but the cell $L$ at the bottom-left corner and the cell $R$ at the top-right corner are not colored black. A knight moves from $L$ to $R$, and every possible path of the knight contains at least one black cell. For which $n>3$ can we assert that there exist three cells connected diagonally, at least two of which are colored black?
4. $n=3 k+1(k=1,2, \cdots)$.

Label the rows from bottom to top as $1,2, \cdots, n$; label the columns from left to right similarly. If the column number and row number of a certain cell are $a$ and $b$ respectively, then the cell is denoted as $(a, b)$.

First, we prove: when $n=3 k, 3 k+2$, it is not always possible to find a cell that meets the requirements.

When $n=3 k$, color all cells in rows whose numbers are congruent to 2 modulo 3 black.

It is not difficult to see that, if the knight starts from cell $L$ and does not pass through any black cell, it can only jump to the cells marked with “ $\times$ ” in Figure 3. Therefore, it cannot reach cell $R$ (Translator's note: This indicates that every possible path from $L$ to $R$ must include a black cell). However, in any three cells connected along a diagonal, there is no more than one black cell.

When $n=3 k+2$, we can similarly color all cells in rows whose numbers are multiples of 3 black (as shown in Figure 4).

Next, we prove: when $n=3 k+1$, it is always possible to find a cell that meets the requirements.
We use induction on $k$.
When $k=1$, the conclusion is obvious. If one of the cells $(2,3)$ and $(3,2)$ is not black, the knight can pass through it to reach the destination (thus, there exists a path that does not require passing through any black cell, which contradicts the problem statement, so we know that cells $(2,3)$ and $(3,2)$ must both be black).
Inductive step.
Assume $k>1$. Suppose the conclusion holds for smaller $k$.

Assume that in any three cells connected along a diagonal, at most one cell is black, but the knight cannot jump from $L$ to $R$. For cell $(a, b)$, let
$$
\left(a^{\prime}, b^{\prime}\right)=(a+3 k-3, b+3 k-3) \text {. }
$$

Thus, if cells $(a, b)$ and $\left(a^{\prime}, b^{\prime}\right)$ are both on the board and not black, according to the inductive hypothesis, the knight can jump from cell $(a, b)$ to $\left(a^{\prime}, b^{\prime}\right)$ (Translator's note: These two cells are equivalent to $L$ and $R$ in an $(n-1) \times (n-1)$ board).

One of the cells $(2,3)$ and $(3,2)$ is not black. Without loss of generality, assume cell $(3,2)$ is not black. If cell $\left(3^{\prime}, 2^{\prime}\right)$ is not black, the knight can follow the path
$$
L \rightarrow(3,2) \rightarrow \cdots \rightarrow\left(3^{\prime}, 2^{\prime}\right) \rightarrow R .
$$

This is impossible. This indicates that cell $\left(3^{\prime}, 2^{\prime}\right)$ is black. According to the assumption, cell $\left(2^{\prime}, 3^{\prime}\right)$ is not black. By similar reasoning, cell $(2,3)$ must be black. Similarly, cell $(4,4)$ must be black, otherwise, there exists a path
$$
L \rightarrow(3,2) \rightarrow(4,4) \rightarrow \cdots \rightarrow\left(4^{\prime}, 4^{\prime}\right)=R \text {. }
$$

Similarly, cell $\left(1^{\prime}, 1^{\prime}\right)$ must be black (as shown in Figure 5).
According to the assumption, the cells marked with “ $\odot$ ” in Figure 5 are all not black. Therefore, the knight can jump from $L$ to cell $(4,1)$, and from there (according to the inductive hypothesis) it can reach cell $\left(4^{\prime}, 1^{\prime}\right)$, and from there to $R$. This contradiction completes the inductive step.