if $a_{ij}=\pm 1$,where $1\le i,j\le 2n$,and if we let two vectors $$X_{u}=(a_{u1},a_{u2},\cdots,a_{u2n}),1\le u\le 2n$$ and $$Y_{v}=(a_{1v},a_{2v},\cdots,a_{2n v}),1\le v\le 2n$$ such for any $1\le i<j\le 2n$ we have $$X_{i}\cdot X_{j}=0$$ show that $$Y_{i}\cdot Y_{j}=0,\forall 1\le i<j\le 2n$$
it seems that the problem is quite simple and straightforward. But how is it strictly proved?
Define the $2n \times 2n$ matrix $A = (a_{ij})$. Note that the scalar product $X_{i}\cdot X_{j}$ is $(i,j)$-th entry of $A A^{T}$, because $$(AA^{T})_{ij} = \sum_{k=1}^{2n}a_{ik}a_{jk}.$$ Now, every such entry must be zero, because $X_{i}\cdot X_{j} = 0$ for every pair of indices $1 \le i,j \le 2n$, so $AA^{T} =0$. Finally, notice that $A^{T}A = (A A^{T})^{T}$.