$2n \times 2n$ matrices with entries in $\{1, 0, -1\}$ with exactly $n$ zeroes in each row and each column with orthogonal rows and orthogonal columns I am interested in answering the following question:
Question
For a given $n$, does there exist a $2n \times 2n$ matrix with entries in $\{1, 0, -1\}$ having orthogonal rows and columns with exactly $n$ zeroes in each row and column?
Conjectures

*

*For $n=2^k$, $k\ge0$ such a matrix always exists.


*For $n=3$ such a matrix does not exist.


*For $n=5$ such a matrix exists, for example:
$$
\left(
\begin{array}{cccccccccc}
 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 & 0 \\
 1 & 0 & 0 & -1 & 0 & 0 & -1 & 1 & 1 & 0 \\
 0 & -1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\
 1 & 0 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -1 \\
 1 & -1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 0 & 0 & -1 & 1 & -1 & 0 & -1 & 0 & 1 \\
 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 & 1 & -1 \\
 0 & -1 & 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \\
 0 & 0 & 0 & -1 & 1 & 1 & 0 & 0 & -1 & -1 \\
\end{array}
\right)
$$


*For $n=7$ such a matrix does not exist.
 A: Your first conjecture was proven by Nate in the comments.
Your second conjecture is also true - there is no such matrix for $n=3$. If we just look at which entries are nonzero in each row, because any two rows are orthogonal, they must share an even number of nonzero entries, i.e. either share $0$ entries or $2$ entries. Only $3$ other rows can share $2$ nonzero entries with a given row, since otherwise there'd be more than $3$ nonzero entries in a column by the pidgeonhole principle, so $2$ rows must share $0$ nonzero entries, but then they share $3$ entries with each other, contradiction.
A: $\def\Id{\text{Id}}$A Hadamard matrix is an $n \times n$ matrix with entries in $\{-1, 1 \}$ with $H H^T = H^T H = n \Id$. What you want is a matrix $X$ with entries in $\{-1, 0, 1 \}$ with $XX^T = X^T X = (n/2) \Id$.
I will show that, if there is an $n \times n$ Hadamard matrix $H$, then there is also an $n \times n$ matrix $X$ obeying your condition. The proof is simple: Let $S = \left[ \begin{smallmatrix} 1/2 & 1/2 \\ 1/2 & - 1/2 \\ \end{smallmatrix} \right]$; note that $S S^T = (1/2) \Id_2$. Let $D = \left[ \begin{smallmatrix} S&&& \\ &S&& \\ &&\ddots& \\ &&&S \\ \end{smallmatrix} \right]$ where there are $n/2$ copies of $S$. We take $X=DH$. Since $H$ has entries in $\{ -1, 1 \}$, the matrix $DH$ has entries in $\{ -1, 0, 1 \}$. We have $(DH)(DH)^T = DHH^TD^T = n DD^T = (n/2)\Id_n$ and similarly $(DH)^T (DH) = (n/2) \Id_n$ as desired.
This is useful because Hadamard matrices are extensively studied. Hadamard's conjecture says that there is always a Hadamard matrix of order $4k$ for any $k$. Hadamard matrices are known to exist for $n$ of the form $q+1$ or $2(q+1)$, where $q$ is a prime power which is $3 \bmod 4$ or $1 \bmod 4$ respectively, and in many other cases; see the above cited Wikipedia article.
That's only half of what you want, because your matrices sometimes exist for $n \equiv 2 \bmod 4$, and Hadamard matrices never do, but it seems like progress.
A: There is no such matrix if $n\equiv 3\pmod 4$.
Suppose otherwise. Each column represents a vector of length $\sqrt n$. Since those vectors are pairwise orthogonal, their sum is a vector whose scalar square is $2n^2$.
On the other hand, the sum of all columns has odd entries, so their squares are all  congruent to $1$ modulo $8$. Hence the sum of those squares is congruent to $2n$ modulo $8$. Hence we should have $2n^2\equiv 2n\pmod 8$, or $8\mid 2n(n-1)$ which does not hold.
