Let $A$ be an antisymmetric matrix of order $(2n+1)\times(2n+1)$, whose non-zero elements are either $1$ or $-1$, and that has no zero rows. I know that
- rank$(A)<2n+1$ since it is not invertible.
- rank$(A)$ must be an even number.
I have a feeling (or rather a wish since it will help me a lot in a different problem I'm working on) that in this particular case it has rank $2n$ but I don't know if it's true \ if it is, how can I prove it.
It is indeed the case that we must have $\operatorname{rank}(A) = 2n$.
As you have noted, $A$ cannot be invertible, so $\operatorname{rank}(A) \leq 2n$. To see that $\operatorname{rank}(A) \geq 2n$, this is the case, it suffices to note that the upper-left $(2n)\times(2n)$ submatrix is a square matrix of even size whose diagonal entries are non-zero and whose non-diagonal entries are either $1$ or $-1$. From this, it follows that this submatrix has a non-zero determinant.