We conjecture that $m=2n$. At first let us show that the condition cant be satisfied by $2n-1$ points. Just check the case where all the $2n-1$ points are along two adjacent sides of the grid. Now, we will show that $2n$ points always satisfy the condition. The grid has $n$ rows and $n$ columns. So, there are $n-1$ possible distance of line by two points of $m$,parallel to the rows. There are always atleast $n$ lines that are made by two points of $m$ and parallel to the rows. Now, the pigeon hole principal. there are two lines of same length. Connect them and we get our parallelogram. What if the two same length lines are on a same line? Let there be $k$ points of $m$ on a single line, there are always $k-1$ different length line possible there. And we are done.

First we'll show that $m=2n-1$ isn't enough. We do that by choosing the whole first row and one of the diagonals. All the points are now on $2$ lines that are not parallel so there are no parallelograms. Now we show by pigeonhole principle that $m=2m$ is a solution. Let's put the grid in a coordinate plane, in a way that the middle of the bottom left square is at $(0,0)$, and the length of square's side is $1$. Now we count the number of different vectors defined by the midpoints of squares. The coefficient next to $\i$ is any whole number between $-(n-1)$ and $n-1$, so there are $2n-1$ possible ways to choose. The same applies for coefficients next to $\j$, so the number of different vectors is $(2n-1)^2$. Now let's see how many vectors we defined with any chosen $2n$ points. The number is $2n(2n-1)$, so by PHP there are some vectors that are equal. The problem occurs if those vectors are collinear, but the number of vectors defined by $2n$ points minus the number of different vectors in total is $2n-1$. There are at most $2n-2$ vectors on each line.

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Excuse my bad