 Solution

Use the standard basis $e_1,\dots ,e_n$. For the smallest possible rank, set $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and set $\beta e_j=-e_j$ for $j\le s$ and zero otherwise. Then the maps have ranks $r,s$, while their sum has nonzero diagonal entries only for $r<j\le s$. HenceFor full rank when $r+s\ge n$, instead take $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and $\beta e_j=e_j$ for $j\ge n-s+1$ and zero otherwise. Their supports cover every basis index because $n-s+1\le r+1$. The sum has diagonal entries one where just one map is nonzero and two where both are nonzero. None vanish over $\mathbb{R}$, so these linear maps have ranks $r,s$ and