1. Assume that there are the invertible matrices $X, Y, Z \in M_n(\mathbb{R})$ such that $AX + YB = AZB$. Then $\text{rank}(A) = \text{rank}(AX) = \text{rank}((AZ - Y)B) \le \text{rank}(B)$. Similarly, $\text{rank}(B) \le \text{rank}(A)$. Therefore, $\text{rank}(A) = \text{rank}(B)$.

2. Assume $\text{rank}(A) = \text{rank}(B) = r$. Then there are the invertible matrices $T, U, V, W \in M_n(\mathbb{R})$ such that
$$TAU = \begin{pmatrix} I_r & O_{r,n-r} \\ O_{n-r,r} & O_{n-r} \end{pmatrix} = VBW.$$
Thus $A(UW^{-1}) = (T^{-1}V)B$. We can choose $\lambda \in \mathbb{R}$ such that $\det(B - \lambda(UW^{-1})) \ne 0$ and $\det(A + \lambda(T^{-1}V)) \ne 0$. Define the invertible matrices $X = B - \lambda(UW^{-1})$ and $Y = A + \lambda(T^{-1}V)$. We obtain
$$AX + YB = 2AB - \lambda A (UW^{-1}) + \lambda (T^{-1}V) B = AZB,$$
where $Z = 2I_n$.