On the $n\times n$ checker board, several cells were marked in such a way that lower left ($L$) and upper right($R$) cells are not marked and that for any knight-tour from $L$ to $R$, there is at least one marked cell. For which $n>3$, is it possible that there always exists three consective cells going through diagonal for which at least two of them are marked?
1. **Define the problem and notation:**
   - We have an \( n \times n \) checkerboard.
   - The lower left cell \( L \) and the upper right cell \( R \) are not marked.
   - For any knight-tour from \( L \) to \( R \), there is at least one marked cell.
   - We need to determine for which \( n > 3 \) it is possible that there always exist three consecutive cells along a diagonal for which at least two of them are marked.

2. **Identify the answer:**
   - The answer is \( \boxed{n \in 3\mathbb{N} + 1} \).

3. **Prove that \( n = 3k + 1 \) works:**
   - Define a "bad" coloring as one where no three consecutive squares along a diagonal have more than one black square.
   - Define a "white" knight tour as one that does not pass through any black squares.

4. **Claim 1:**
   - For \( n = 3k + 1 \), in a bad coloring of an \( n \times n \) checkerboard, there exists a white knight tour.

5. **Proof by strong induction on \( k \):**
   - **Base case:** For \( n = 4 \), it is trivial.
   - **Inductive step:** Assume the claim holds for \( n = 3m + 1 \) for all \( m < k \). Consider \( n = 3k + 1 \).

6. **Assign coordinates:**
   - Let \( L = (1,1) \) and \( R = (n,n) \).
   - Define the set \( A = \{(a,a) \mid a = 3i + 1, 1 \le i \le k-1\} \).

7. **Case (i): \( A \) contains a white cell:**
   - Let the white cell be \( (3j + 1, 3j + 1) \).
   - Consider the sub-boards determined by the corners \( C_1 = \{L, (1, 3j + 1), (3j + 1, 1), (3j + 1, 3j + 1)\} \) and \( C_2 = \{(3j + 1, 3j + 1), (3j + 1, n), (n, 3j + 1), R\} \).
   - By induction hypothesis, we can move the knight from \( L \) to \( (3j + 1, 3j + 1) \) and then from \( (3j + 1, 3j + 1) \) to \( R \), completing a white knight tour.

8. **Case (ii): All cells in \( A \) are black:**
   - The board is symmetric along the diagonal.
   - Assume \( (3,2) \) is white (since both \( (2,3) \) and \( (3,2) \) cannot be black).

9. **Subcase (a): \( (3k, 3k-1) \) is white:**
   - Move the knight from \( L \) to \( (3,2) \).
   - Consider the square determined by \( C = \{(3,2), (3k,2), (3k,3k-1), (3,3k-1)\} \).
   - Using induction, move the knight from \( (3,2) \) to \( (3k,3k-1) \) and then directly to \( R \).

10. **Subcase (b): \( (3k, 3k-1) \) is black:**
    - Move the knight to \( (3,2) \).
    - \( (5,3) \) is white since \( (4,4) \) is black.
    - Move the knight to \( (5,3) \).
    - Continue this process using smaller \( 4 \times 4 \) checkerboards until reaching \( (3k-1, 3k-3) \).
    - Use the sequence of moves to reach \( R \).

11. **Conclusion for \( n = 3k + 1 \):**
    - The induction is complete, proving that \( n = 3k + 1 \) works.

12. **Prove that other \( n \) do not work:**
    - **Case \( n = 3k \):**
      - Color the whole \( (3h+2)^{\text{th}} \) row black for \( h = 0, 1, 2, \ldots, k-1 \).
      - Example for \( 9 \times 9 \) board shows no valid knight tour.

    - **Case \( n = 3k + 2 \):**
      - Color the whole \( (3h)^{\text{th}} \) row black for \( h = 1, 2, \ldots, k \).
      - Example for \( 8 \times 8 \) board shows no valid knight tour.

13. **Conclusion:**
    - For \( n = 3k \) or \( n = 3k + 2 \), it is not possible to ensure a valid knight tour.