Let $H_1, H_2$ be Hilbert spaces. Let $x \in M_n(B(H_1, H_2))$. Then define $x^* \in M_n(B(H_2, H_1))$ in the usual way, i.e. $(x^*)_{i,j}:= x_{j,i}^* \in B(H_2, H_1)$. There is also an obvious way to multiply the matrices $x^*$ and $x$ and form the product $$x^* x \in M_n(B(H_1)) = M_n(B(H_1, H_1))$$
Is it true that $x^* x$ is a positive matrix of the $C^*$-algebra $M_n(B(H))= B(H^n)$? Maybe I can calculate $$\langle x^*x \xi, \xi\rangle_{H^n}$$ and sho that is $\geq 0$? What is the easiest way to show this?
Computing $\langle x^*x \xi, \xi\rangle_{H^n}$ seems to be the most reasonable approach. In particular, the definition of the adjoint means that we have $$ \langle x^*x \xi, \xi\rangle_{H_1^n} = \langle x \xi, x\xi\rangle_{H_2^n} = \|x \xi\|_{H_2^n}^2 \geq 0, $$ so that $x^*x$ is indeed a positive map.
Note: this assumes that relative to your "usual" definition of the adjoint, we have $$ \langle x^*\xi_2,\xi_1 \rangle_{H_1^n} = \langle \xi_2, x \xi_1 \rangle_{H_2^n}, $$ where $\xi_1 \in H_1^n$ and $\xi_2 \in H_2^n$. This is indeed the case, but perhaps you should prove this in detail.