Let $n$ be a natural number divisible by $3$. We have a $n \times n$ table and each square is colored either black or white. Suppose that for all $m \times m$ sub-tables from the table ($m > 1$), the number of black squares is not more than white squares. Find the maximum number of black squares.
1. **Understanding the Problem:**
   We need to find the maximum number of black squares in an \( n \times n \) table such that for any \( m \times m \) sub-table (where \( m > 1 \)), the number of black squares is not more than the number of white squares. Given that \( n \) is divisible by 3, we can write \( n = 3k \) for some integer \( k \).

2. **Construction of the Table:**
   We color the cells with coordinates \((a, b)\) black if and only if \((a, b)\) are \((1,2), (2,1), (3,2), (2,3)\) modulo 3. This means:
   - A cell \((a, b)\) is black if \( a \equiv 1 \pmod{3} \) and \( b \equiv 2 \pmod{3} \)
   - A cell \((a, b)\) is black if \( a \equiv 2 \pmod{3} \) and \( b \equiv 1 \pmod{3} \)
   - A cell \((a, b)\) is black if \( a \equiv 3 \pmod{3} \) and \( b \equiv 2 \pmod{3} \)
   - A cell \((a, b)\) is black if \( a \equiv 2 \pmod{3} \) and \( b \equiv 3 \pmod{3} \)

3. **Counting the Black Squares:**
   Each \( 3 \times 3 \) block in the table will have exactly 4 black squares. Since the table is \( n \times n \) and \( n = 3k \), there are \( k^2 \) such \( 3 \times 3 \) blocks in the table. Therefore, the total number of black squares is:
   \[
   4 \times k^2 = 4 \times \left(\frac{n}{3}\right)^2 = \frac{4n^2}{9}
   \]

4. **Verification:**
   We need to verify that any \( m \times m \) sub-table (where \( m > 1 \)) has at most half of its cells black. Consider any \( 2 \times m \) or \( 3 \times m \) rectangle:
   - For \( 2 \times m \) rectangles, since \( m \) can be written as \( 2a + 3b \) for non-negative integers \( a \) and \( b \), the number of black squares will be at most half of the total cells.
   - For \( 3 \times m \) rectangles, the same logic applies, ensuring that the number of black squares is at most half of the total cells.

   This construction ensures that the condition is satisfied for all \( m \times m \) sub-tables where \( m > 1 \).

The final answer is \(\boxed{\frac{4n^2}{9}}\)