For $n>1$ consider an $n\times n$ chessboard and place identical pieces at the centers of different squares.

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[*] Show that no matter how $2n$ identical pieces are placed on the board, that one can always find $4$ pieces among them that are the vertices of a parallelogram.
[*] Show that there is a way to place $(2n-1)$ identical chess pieces so that no $4$ of them are the vertices of a parallelogram.
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### Part (i)
1. **Statement of the problem**: We need to show that no matter how \(2n\) identical pieces are placed on an \(n \times n\) chessboard, one can always find 4 pieces among them that are the vertices of a parallelogram.

2. **Pigeonhole Principle**: Consider the \(n \times n\) chessboard. Each piece is placed at the center of a square. There are \(n^2\) squares on the board. We place \(2n\) pieces on the board.

3. **Pairs of Rows and Columns**: There are \(\binom{n}{2} = \frac{n(n-1)}{2}\) ways to choose 2 rows and \(\binom{n}{2} = \frac{n(n-1)}{2}\) ways to choose 2 columns. Each pair of rows and each pair of columns can form a parallelogram if there are pieces at the intersections of these rows and columns.

4. **Counting Pairs**: The total number of pairs of rows and columns is \(\left(\frac{n(n-1)}{2}\right)^2 = \frac{n^2(n-1)^2}{4}\).

5. **Pigeonhole Application**: Since we have \(2n\) pieces, by the pigeonhole principle, there must be at least one pair of rows and one pair of columns such that there are at least 2 pieces in each of these rows and columns. This guarantees the formation of a parallelogram.

6. **Conclusion**: Therefore, no matter how \(2n\) pieces are placed on the board, one can always find 4 pieces among them that are the vertices of a parallelogram.

\(\blacksquare\)

### Part (ii)
1. **Statement of the problem**: We need to show that there is a way to place \((2n-1)\) identical chess pieces so that no 4 of them are the vertices of a parallelogram.

2. **Construction of Placement**: Consider the points \(\{(1,x), (x,1) : x \in [n]\}\) on the \(n \times n\) grid. This means we place pieces at \((1,1), (1,2), \ldots, (1,n)\) and \((2,1), (3,1), \ldots, (n,1)\).

3. **Verification**: We need to verify that no 4 of these points form a parallelogram. Suppose there are 4 points that form a parallelogram. Then, these points must lie on two distinct rows and two distinct columns.

4. **Counting Distances**: Suppose there are \(c_i\) chosen points in row \(i\). Within each row, there are at least \(c_i - 1\) distinct distances. If there was the same distance in two rows, we would have a parallelogram. The only possible distances are \(1, \ldots, n-1\).

5. **Summing Distances**: Thus, we have:
   \[
   \sum_{i=1}^n (c_i - 1) \leq n - 1
   \]
   This implies:
   \[
   \sum_{i=1}^n c_i \leq 2n - 1
   \]

6. **Conclusion**: Therefore, it is possible to place \((2n-1)\) pieces on the board such that no 4 of them form the vertices of a parallelogram.

\(\blacksquare\)

The final answer is \( \boxed{ 2n-1 } \)