Prove that, at least one of the matrices $A+B$ and $A-B$ has to be singular

(1) If $X$ is real $n\times n$ skew symmetric matrix, i.e., $X+X^T=0$, and if $T$ means transpose, then $$ {\rm det}\ X={\rm det}\ -X^T={\rm det}\ (-I)X={\rm det}\ X (-1)^n $$ where $I$ is an identity. So if $n$ is odd then ${\rm det}\ X=0$

(2) Assume that $A,\ B$ are orthogonal Then since $AA^T=I=BB^T$, $$ (A+B)(A-B)^T= BA^T - AB^T $$

Note that if $Y=BA^T$, then $(A+B)(A-B)^T=Y-Y^T$ is skew symmetric. So ${\rm det}\ (A+B)(A-B)^T =0$. So at leat one of $A+B,\ A-B$ is singular.

You can write $$ A\pm B=A(\mathbf 1 \pm A^T B)=A(\mathbf 1 \pm Q), \quad Q\textrm{ orthogonal} $$ Then $A\pm B $ is singular iff $\mathbf 1 \pm Q $ is singular. But $Q$ has always an eigenvector $v$ with eigenvalue $\pm 1$ when $n$ is odd, therefore $\mathbf 1 -Q $ or $\mathbf 1 + Q$ has to be singular.