There are given $111$ coins and a $n\times n$ table divided into unit cells. This coins are placed inside the unit cells (one unit cell may contain one coin, many coins, or may be empty), such that the difference between the number of coins from two neighbouring cells (that have a common edge) is $1$. Find the maximal $n$ for this to be possible.
1. **Understanding the Problem:**
   We need to place 111 coins in an \( n \times n \) grid such that the difference between the number of coins in any two neighboring cells (cells sharing a common edge) is exactly 1. We aim to find the maximum possible value of \( n \).

2. **Constructing the Grid for \( n = 13 \):**
   Let's consider \( n = 13 \). We can construct a grid where the coins are placed in a specific pattern:
   - Place 1 coin in each cell of the odd-indexed rows (1st, 3rd, 5th, ..., 13th).
   - Place 0 coins in each cell of the even-indexed rows (2nd, 4th, 6th, ..., 12th).

   This pattern ensures that the difference between the number of coins in any two neighboring cells is 1.

3. **Counting the Coins:**
   - There are 7 odd-indexed rows, each with 13 cells, and each cell contains 1 coin. Thus, the total number of coins in the odd-indexed rows is:
     \[
     7 \times 13 = 91
     \]
   - There are 6 even-indexed rows, each with 13 cells, and each cell contains 0 coins. Thus, the total number of coins in the even-indexed rows is:
     \[
     6 \times 13 = 0
     \]
   - The total number of coins in the grid is:
     \[
     91 + 0 = 91
     \]

4. **Adjusting the Number of Coins:**
   Since we need a total of 111 coins, we need to add 20 more coins to the grid. We can achieve this by increasing the number of coins in some of the cells that currently contain 0 coins to 2 coins. Specifically, we can select 20 cells that currently contain 0 coins and place 2 coins in each of these cells.

5. **Ensuring the Difference Condition:**
   By placing 2 coins in 20 cells that originally contained 0 coins, we ensure that the difference between the number of coins in any two neighboring cells remains 1. This is because the cells that originally contained 0 coins are surrounded by cells that contain 1 coin, and increasing the number of coins in these cells to 2 maintains the difference condition.

6. **Conclusion:**
   Therefore, it is possible to place 111 coins in a \( 13 \times 13 \) grid while satisfying the given condition. We need to verify if \( n = 14 \) or higher is possible.

7. **Verification for \( n = 14 \):**
   - For \( n = 14 \), the total number of cells is \( 14 \times 14 = 196 \).
   - If we follow a similar pattern, we would place 1 coin in each cell of the odd-indexed rows and 0 coins in each cell of the even-indexed rows.
   - There are 7 odd-indexed rows, each with 14 cells, and each cell contains 1 coin. Thus, the total number of coins in the odd-indexed rows is:
     \[
     7 \times 14 = 98
     \]
   - There are 7 even-indexed rows, each with 14 cells, and each cell contains 0 coins. Thus, the total number of coins in the even-indexed rows is:
     \[
     7 \times 14 = 0
     \]
   - The total number of coins in the grid is:
     \[
     98 + 0 = 98
     \]
   - To reach 111 coins, we need to add 13 more coins. We can place 2 coins in 13 cells that originally contained 0 coins.

8. **Ensuring the Difference Condition for \( n = 14 \):**
   - By placing 2 coins in 13 cells that originally contained 0 coins, we ensure that the difference between the number of coins in any two neighboring cells remains 1.

9. **Conclusion for \( n = 14 \):**
   - Therefore, it is possible to place 111 coins in a \( 14 \times 14 \) grid while satisfying the given condition.

10. **Verification for \( n = 15 \):**
    - For \( n = 15 \), the total number of cells is \( 15 \times 15 = 225 \).
    - If we follow a similar pattern, we would place 1 coin in each cell of the odd-indexed rows and 0 coins in each cell of the even-indexed rows.
    - There are 8 odd-indexed rows, each with 15 cells, and each cell contains 1 coin. Thus, the total number of coins in the odd-indexed rows is:
      \[
      8 \times 15 = 120
      \]
    - There are 7 even-indexed rows, each with 15 cells, and each cell contains 0 coins. Thus, the total number of coins in the even-indexed rows is:
      \[
      7 \times 15 = 0
      \]
    - The total number of coins in the grid is:
      \[
      120 + 0 = 120
      \]
    - Since 120 is greater than 111, we can place 111 coins in a \( 15 \times 15 \) grid while satisfying the given condition.

The final answer is \( \boxed{ n = 15 } \)