Let us assume that I have a matrix $D\in\Re^{2N\times 2N}$ with the following structure: $$ D=\begin{bmatrix} 0 & A \\ A^T & 0 \\ \end{bmatrix} \quad $$
where $A \in\Re^{N\times N}$. Is it possible to find an $A$ such that $D > 0$? I would like $A$ to be orthogonal.
Here are two reasons for which it is not possible :
a) Such a matrix cannot fullfill Sylvester criterion for positive definiteness : all principal minors have to be $>0$.
b) The eigenvalues of $D$ come by pairs of opposite values (i.e., if $\lambda$ is an eigenvalue of $D$, so is $-\lambda$) meaning that there are negative eigenvalues which is not possible for a psd matrix.
More precisely : the spectrum of $D$ is $S \cup (-S)$ where $S$ is the set of (positive !) singular values of $A$.
Let us prove it. Let $A=USV^T$ be the Singular Values Decomposition of $A$. One can write :
$$D=\begin{pmatrix}0&A\\A^T&0\end{pmatrix}=\underbrace{\begin{pmatrix}U&0\\0&V\end{pmatrix}}_{P}\underbrace{\begin{pmatrix}0&S\\S&0\end{pmatrix}}_{\Delta}\underbrace{\begin{pmatrix}U^T&0\\0&V^T\end{pmatrix}}_{P^T=P^{-1}}$$
$D$ and $\Delta$, being similar, have the same spectrum. The spectrum of $\Delta$ is given by $$\det(\Delta-\lambda I_{2n})=0$$ which in fact is equal to
$$\det(\lambda^2I_n-S^2)=\prod_{k=1}^n(\lambda^2-\sigma_k^2)$$
(due to rule $\det\begin{pmatrix}A&B\\C&D\end{pmatrix}=\det(AD-BC)$ valid when, for example, $C$ and $D$ commute ; see theorem 3 in this detailed document)
Very similar proofs here and here.