Let $n$ be an even integers. Let $r\in \mathbb R^n$ and $e=[1,1,\dots,1]^T$. If $$A = re^T - er^T,$$ then $A\in \mathbb{R}^{n\times n}$ is of rank 2 and skew-symmetric, i.e., $$A = -A^T.$$
This does not represent all rank skew-symmetric matrices, but a useful subset which appears in ranking problems. Clearly there are no more than $n$ degrees of freedom in choosing $A$.
Does there exist a similar method of generating an arbitrary (even) rank skew-symmetric matrix? Skew-symmetric matrices of any rank in general have $\frac{n(n-1)}{2}$ degrees of freedom, representing $\binom{n}{2}$ row/column pairs. I'm looking for skew-symmetric matrices that somehow represent requiring much fewer comparisons than this such that the space of possible $A$ has fewer than $\binom{n}{2}$ degrees of freedom.
Let $A= -A^T$. Let $e_i$ denote the $i-th$ unit vector. Then it holds $$ A = \sum_{i=2}^n\sum_{j=1}^{i-1} a_{ij}( e_ie_j^T-e_je_i^T). $$ In that way you represent any skew-symmetric matrix as a sum of rank-2 matrices.
By truncating the sum, you obtain matrices with a specified even rank. Is this what you are after?