Why are $\{0,1\}$-matrices almost-MDS only when n is 2, 3, or 4?
From the article on page 5;

Theorem 1 Let $A:{GF(2^m)}^n \to {GF(2^m)}^n$ be an $n\times n \{0,1\}$-matrix over $GF(2^m)$. Then the branch number of $A$ is at most $\frac{2n+4}{3}$.

Let $A$ be the almost-MDS matrix. So its branch number is $n$. Using above theorem we have
$$n \leq \frac{2n+4}{3} $$
So $n=2,3$ or $4$.
The size of matrix is a tool for finding upper bound of branch number. The branch number of $n\times n$ matrix is at most $n+1$ and equality occur, when matrix be MDS.
Also matrix $A$ is MDS if determinant of all its sub matrices be not $0$.