Arnaldo and Bernardo play a Super Naval Battle. Each has a board $n \times n$. Arnaldo puts boats on his board (at least one but not known how many). Each boat occupies the $n$ houses of a line or a column and the boats they can not overlap or have a common side. Bernardo marks $m$ houses (representing shots) on your board. After Bernardo marked the houses, Arnaldo says which of them correspond to positions occupied by ships. Bernardo wins, and then discovers the positions of all Arnaldo's boats. Determine the lowest value of $m$ for which Bernardo can guarantee his victory.
To determine the lowest value of \( m \) for which Bernardo can guarantee his victory, we need to ensure that Bernardo's shots can always reveal the positions of all of Arnaldo's boats. Let's analyze the problem step-by-step.

1. **Case \( n \equiv 0 \mod 3 \):**
   - Divide the \( n \times n \) board into \( \frac{n}{3} \times \frac{n}{3} \) smaller \( 3 \times 3 \) squares.
   - For each \( 3 \times 3 \) square, Bernardo will choose all squares in the first row and all squares in the last column, except the square at the top right. This results in 4 chosen cells per \( 3 \times 3 \) square.
   - Since there are \( \frac{n}{3} \times \frac{n}{3} = \left(\frac{n}{3}\right)^2 \) such squares, the total number of chosen cells is:
     \[
     m = 4 \times \left(\frac{n}{3}\right)^2 = \frac{4n^2}{9}
     \]
   - This ensures that each boat, whether horizontal or vertical, will intersect with at least one chosen cell, allowing Bernardo to determine the orientation and position of the boats.

2. **Case \( n \equiv 1 \mod 3 \):**
   - Similar to the previous case, divide the board into \( 3 \times 3 \) squares, but there will be an extra row and column.
   - Use the same strategy for the \( 3 \times 3 \) squares, and additionally, choose cells in the extra row and column to ensure coverage.
   - The total number of chosen cells will be:
     \[
     m = \left\lceil \frac{4n}{3} \right\rceil + 1
     \]

3. **Case \( n \equiv 2 \mod 3 \):**
   - Again, divide the board into \( 3 \times 3 \) squares, but there will be two extra rows and columns.
   - Use the same strategy for the \( 3 \times 3 \) squares, and additionally, choose cells in the extra rows and columns to ensure coverage.
   - The total number of chosen cells will be:
     \[
     m = \left\lceil \frac{4n}{3} \right\rceil + 2
     \]

4. **Minimality Argument:**
   - Suppose \( m < \left\lceil \frac{4n}{3} \right\rceil \). Divide the board into \( 3 \times n \) horizontal stripes. By the pigeonhole principle, there is at least one stripe with at most 3 marked houses on it.
   - Similarly, for \( n \times 3 \) vertical stripes, there is at least one stripe with at most 3 marked houses on it.
   - Consider the union of these two chosen stripes. It is impossible for Bernardo to distinguish between two vertical boats inside this union and two horizontal boats inside this union with fewer than \( \left\lceil \frac{4n}{3} \right\rceil \) shots.

Thus, the minimal value of \( m \) for which Bernardo can guarantee his victory is:
\[
m = \left\lceil \frac{4n}{3} \right\rceil + c
\]
where \( c \) is the remainder of \( n \mod 3 \).

The final answer is \( \boxed{ \left\lceil \frac{4n}{3} \right\rceil + c } \).