Consider the matrix family $M_n=\begin{bmatrix} 0&1&1&\dots&1&1\\ -1&0&1&\dots&1&1\\ -1&-1&0&\dots&1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ -1&-1&-1&\dots&0&1\\ -1&-1&-1&\dots&-1&0 \end{bmatrix}_{n\times n}$ where $M_{i,i}=0$ and $M_{i,j}=-M_{j,i}=1$ if $j>i$.
Why is the rank of the matrix $M_n$ in family always $2\lfloor\frac n2\rfloor$?
First, we have the following claim, which is not hard to prove:
Claim. The span of vectors $v_1, \ldots, v_m$ have the same dimension as the span of vectors $v_1-v_2, v_2-v_3, \ldots, v_{m-1}-v_m,v_m+v_1$.
We apply the following claim to the columns of the matrix. Hence, it suffices to find the rank of the following matrix:
$$A=\begin{pmatrix} -1&0&0&\dots&0&0&1\\ -1&-1&0&\dots&0&0&0\\ 0&-1&-1&\dots&0&0&0\\ 0&0& -1 & \dots &0&0&0 \\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\dots&-1&-1&0\\ 0&0&0&\dots&0&-1&-1 \end{pmatrix}$$ Denote the columns of $A$ from left to right as $v_1, \ldots, v_n$. Let see if these columns are linearly independent or not. Consider $\alpha_i \in \mathbb{R}$ such that $\sum_{i=1}^n \alpha_iv_i=0$. For each row $i =2, \ldots, n$ we find $\alpha_{i-1}+\alpha_i=0$. From these equations, we obtain that $\alpha_1=\alpha_3= \alpha_5= \ldots= \alpha_{2\left \lfloor (n-1)/2 \right \rfloor +1}$ and $\alpha_2=\alpha_4= \ldots = \alpha_{2 \left \lfloor n/2\right \rfloor}$. We have not consider the first row, which gives $\alpha_1-\alpha_n=0$.
If $n$ is even and from $\alpha_1=\alpha_{n}$, we obtain $\alpha_i=0$ for all $1 \le i \le n$. This follows $v_1, \ldots, v_n$ is linearly independent, which gives the dimension of $n$.
If $n$ is odd, then surely the columns are not linearly independent, as we can choose nonzero $\alpha_i$'s as follow: $\alpha_1=\alpha_3= \ldots= \alpha_n=1$ and $\alpha_2=\alpha_4= \ldots = \alpha_{n-1}=-1$. To show that dimension of the columns $v_1, \ldots, v_n$ is $n-1$, we take out any column, WLOG say $v_n$, and then show that $v_1, \ldots, v_{n-1}$ are linearly independent. This can be done similarly as above by writing simple equation for each row.
Thus, our final answer is indeed $2 \left \lfloor \frac n2 \right \rfloor$.