Problem 1430

### 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 $\boxed{ 2n-1 }$