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Yes, it is possible; pick a nonzero element of $\mathbf{x}$, which must exist since $\mathbf{x}\neq 0$; let's say this nonzero element is $x_j$. Then define $$\big(\mathbf{M}_{\mathbf{x}}\bigr)_{nm}=-\delta_{nm}(1-\delta_{nj})+\delta_{nm}\delta_{nj}\left(x_{j}^{-2}+x_{j}^{-2}\sum_{k\neq j}x_k^2\right).$$ Now you have the desired result, $$\sum_{n,m}x_nx_m\big(\mathbf{M}_{\mathbf{x}}\bigr)_{nm}=1>0,$$ while $\mathbf{M}_{\mathbf{x}}$ is itself not positive definite.