Try this with a trapezium instead

We generalize the problem as follows: If counters are placed in the squares of an $m \times n$ grid, where $m \ge 2$ and $n \ge 2$ , then the largest number of counters that can be placed so that no four counters form a non-degenerate parallelogram is $m + n - 1$ .

Proof : Consider an $m \times n$ grid that contains at least $m + n$ counters. We call a row crowded if it contains at least two counters. Let $r$ be the number of crowded rows. There must be at least two crowded rows; otherwise, the number of counters would be at most $n + (m - 1) = m + n - 1$ .

Each of the remaining $m - r$ rows that are not crowded contain at most one counter, so the number of counters among all crowded rows is at least $m + n - (m - r) = n + r$ .

For each crowded row, we record the distance between the leftmost counter and every other counter in the same row. Each such distance is between 1 and $n - 1$ .

Let $a_i$ be the number of counters in the $i^{\text{th}}$ crowded row. Then we record $a_i - 1$ distances for the $i^{\text{th}}$ crowded row, so the total number of distances that we record is $(a_1 - 1) + (a_2 - 1) + \dots + (a_r - 1) = a_1 + a_2 + \dots + a_r - r,$ which is at least $(n + r) - r = n$ .

Hence, by the Pigeonhole Principle, two of the distances that we recorded are equal. The four counters that correspond to these equal distances form a non-degenerate parallelogram.

Furthermore, if we place a counter in each square in the first row and first column, then we obtain a grid with $m + n - 1$ counters, where no four counters form a non-degenerate parallelogram.

For any $1 \le y \le 30$ let $B_y = \{1 \le x \le 20 | (x,y) \in B\}$ and let $C_j \; = \; \{1 \le y \le 30 | |B_y| = j\} \qquad 1 \le j \le 20$ noting that $\sum_{j=1}^{20}|C_j| \le 30 \qquad \sum_{j=1}^{20}j|C_j| = |B|$ For any $2 \le j \le 20$ and $y \in C_j$ , let $B_y = \{x_1,x_2,\ldots,x_j\}$ where $x_1 < x_2 < \cdots < x_j$ . Then $D_y = \{x_k - x_1 | 2 \le k \le j\}$ is a set of $j-1$ distinct numbers between $1$ and $19$ . The sets $D_y$ (for $y \in C_j$ and $2 \le j \le 20$ ) must be disjoint, since otherwise we could find distinct $y_1,y_2$ and distinct $x_{11},x_{12} \in B_{y_1}$ and distinct $x_{21},x_{22} \in B_{y_2}$ with the same nonzero difference $|x_{11} - x_{12}| = |x_{21} - x_{22}|$ in $D_{y_1} \cap D_{y_2}$ , and hence obtain a nondegenerate parallelogram in $B$ (with one horizontal pair of sides). Thus we deduce that $\sum_{j=2}^{20}(j-1)|C_j| \le 19$ and so $|B| = \sum_{j=1}^{20}j|C_j| = \sum_{j=1}^{20}|C_j| + \sum_{j=2}^{20}(j-1)|C_j| \le 30 + 19 = 49$

The set $B = \big\{(1,y) \big| 1 \le y \le 30\big\} \cup \big\{(x,1) \big| 2 \le x \le 20\big\}$ is a concrete example of a set $B$ with $49$ elements.

Note that the set $S$ can be interpreted as a $20 \times 30$ rectangular grid. We shall prove by induction that for all $a,b \in \mathbb{N}_{>1}$ , the largest possible number of lattice points inside any rectangular $a \times b$ grid such that no four of them form a parallelogram is $a+b-1$ . For the base case, we have to show that in a $2 \times 2$ rectangular grid, the largest possible number of such points is $2+2-1= 3$ . This is trivial, since if we choose $4$ points, we get the $2 \times 2$ rectangular grid itself, which is a parallelogram.

We shall now show that if our assertion is true for rectangular $a \times b$ grid and a rectangular $(a-1) \times b$ grid, it will also be true for a rectangular $(a+1) \times b$ grid. Note that if we can prove this, we will be done, since by symmetry our claim will also hold true for a rectangular $a \times (b+1)$ grid, which will complete the induction step on $a$ and $b$ .

We have assumed that for a rectangular $a \times b$ grid, the largest number of such points is $a+b-1$ . Suppose we increase $a$ by $1$ . We have to show that we can incorporate at most $1$ lattice point in our set. On the contrary, assume we can take $2$ more points in our selection. We divide the $(a+1) \times b$ grid into two rectangular grids: the first grid $G_1$ consisting of the first $a$ columns, and the second grid $G_2$ consisting of the last $a$ columns. Note that the dimensions of each these grids is $a \times b$ , and hence can accommodate at most $a+b-1$ lattice points. Also note that $G_1 \cap G_2$ is a $(a-1) \times b$ rectangular grid, and can hence accommodate at most $a+b-2$ lattice points. Thus, atleast $3$ points must lie within either the first column or the last column. Either way, the column with the greater number of points must have at least $\left \lceil \frac{3}{2} \right \rceil = 2$ lattice points. WLOG assume the first column has more points than the last column. It is then easy to see that $G_1$ has more than $a+b-1$ lattice points, a contradiction. This completes our induction step.

It is easy to see that equality holds when we take the $a+b-1$ points in a $\text{L}$ or $\text{I}$ -shaped figure at the edge of the rectangular grid. Our desired answer will then be $20+30-1= \boxed{49}$ .

We can easily exhibit a set with $49$ elements by simply picking points having either $x=1$ or $y=1$ . To show that this is the best possible, suppose $B$ is maximal and has $50$ elements. We can assume that $B$ contains at least one element from each column (otherwise it's easy to add some point from that column obtaining a bigger set). Subtracting these $30$ points (one per column), there must be $20$ more points somewhere.

The idea now is that putting a point in any column will form a line with the point that we have already selected in the previous step. The length of such a line is between $1$ and $19$ . Therefore, by pigeonhole principle, there are two columns containing lines of the same length. But that's the same as saying that the four points from those lines form a parallelogram, in contradiction with $B$ being allowed.

Interesting problem. My first guess is that the answer is 49. Take all points where the x or y coordinate is 1. That gives you a column with 30 points, plus a row with 20, less the one point where they intersect. 30+20-1=49.. Add any other point and you get a rectangle.