For what $n \geq 3$ does there exist an $n \times n$ matrix such that:


* All entries are in $(0, 1)$.
* Each row and column sums to $1$.
* Aside from the rows and columns, no other subsets of the entries sum to $1$.


EDIT: I had a comment about $n = 3$ likely not being possible, but I removed that now after some helpful comments.



Magic squares with specific properties
Such a matrix (integer version) can be constructed from a magic square, say with entries $1,\dots,n^2$, by adding suitable multiples of all $n\times n$ permutation matrices: e.g. take $N=n^2$ and add all $N^iP\_i$, where $P\_1, \dots,P\_{n!}$ are in any order.  

This is such a wasteful construction that the question raised in my comment above might deserve some attention: what are good upper bounds for the common row and column sum of such a matrix? 


