Consider a separable state $\rho$ living in a tensor product space $\mathcal H\otimes\mathcal H'$, with $\mathcal H$ and $\mathcal H'$ of dimensions $D$ and $D'$, respectively. If $\rho$ is separable, then it is by definition possible to write it as a convex combination of (projectors over) separable pure states.
$\newcommand{\ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert}$Because a state is Hermitian and positive by definition, we can trivially always write it in terms of its eigenvectors and eigenvalues as $$ \rho = \sum_{k=1}^{D^{} D^\prime} \lambda_k \ketbra{\psi_k}, \quad p_k\ge0, $$ where $\rho|\psi_k\rangle=\lambda_k|\psi_k\rangle$. However, $|\psi_k\rangle$ will in general be non-separable states.
What I am looking for is the decomposition of $\rho$ in terms of only separable states. For example, a trivial case is $\rho=I/DD'$, which is easily seen to be decomposable as $$\frac{1}{DD'}I=\frac{1}{DD'}\sum_{k=1}^D\sum_{\ell=1}^{D'}\ketbra{k,\ell}.$$ This shows that, to decompose an unknown state $\rho$ in terms of separable states, at least $DD'$ elements are required. Is this number sufficient for any separable $\rho$?
In other words, what I'm looking for is the smallest $M$ such that a representation of the form $$\rho = \sum_{j=1}^M p_j \,\ketbra{\alpha_j}\otimes\ketbra{\beta_j}$$ holds for all separable $\rho$. More formally, this amounts to finding $$\min\left\{M\in\mathbb N\,:\,\,\forall\rho\exists\{p_k\}_k,\{|\alpha_k\rangle\}_k,\{|\beta_k\rangle\}\,:\,\rho=\sum_{j=1}^M p_j \,\ketbra{\alpha_j}\otimes\ketbra{\beta_j}\right\}.$$