Let $H_n$ be an $n×n$ Hadamard matrix and $R_n$ the $n×n$ reverse identity matrix.
The matrix $X= \begin{pmatrix} H_n & R_nH_n \\ H_n & -R_nH_n \end{pmatrix}$ has entries of length $1$ and $$XX^T = 2nI_{2n} + ((nI_n - H_nR_n^*R_nH_n^*) \otimes R_2)$$ which is simply $2nI_{2n}$ so it is a Hadamard matrix. Permute the last $n$ columns with $R_n$ and you have it in the form you give. This gives half-skew-centrosymmetric Hadamard matrices of twice the size of a Hadamard matrix. (It also works for complex Hadamard matrices or even Hadamard matrices over *-rings)
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(The stuff below is not what you asked, but I gave it the trouble of thinking it out so I would regret deleting this):
Complex and real Hadamard matrices may also be constructed by the Paley construction I from finite fields of order $4k+3$. Instead of taking $I+X$ for $X$ the jacobsthal matrix with added $1$-row above and $-1$-column left (zero diagonal), one can replace $I$ by $R_2 H_2 \otimes I$ and also use $H_2 \otimes X$. This is a Hadamard matrix since $$(R_2 H_2 \otimes I + H_2 \otimes X) (R_2 H_2 \otimes I + H_2 \otimes X)^*$$ can be reduced to $$H_2H_2^* \otimes nI + R_2H_2H_2^* \otimes X + H_2H_2^*R_2^* \otimes X^*$$ and of course $H_2H_2^*=I_2$ and $R_2=R_2^*$ commutes with it so this becomes $$2nI_{2n} + R_2 \otimes (X+X^*)$$ and $X^* = -X$ giving the result. One also needs to check the entries have norm $1$, which I leave to the reader.