Given a perfect fractional matching, does there exist a perfect matching with heavy edges? Let $G = (X\cup Y, E)$ be a bipartite graph in which $|X|=|Y|=n$. Suppose $G$ admits a perfect fractional matching, that is - a function assigning a non-negative weight to each edge, such that the sum of weights of edges near each vertex is exactly $1$.
It is known that such a $G$ always admits a perfect matching. One way to prove it is using Hall's marriage theorem: for each subset of $k$ vertices of $X$, the sum of weight near these vertices is $k$, so they must be adjacent to at least $k$ vertices of $Y$. Thus $G$ satisfies Hall's condition.
What is the largest $r(n)$ such that $G$ always admits a perfect matching in which the weight of every edge is at least $r(n)$?
An upper bound on $r(n)$ is $1/n$.  It is given by the complete biartite graph and the fractional matching in which the weight of each edge is $1/n$.
A lower bound on $r(n)$ is $1/n(n-1)$. Proof: remove from $G$ all edges with a weight of less than $1/n(n-1)$. For each vertex $v$, we removed at most $n-1$ edges adjacent to $v$ (since at least one edge must remain). Hence, the weight near $v$ decreased by less than $1/n$, and the remaining weight is more than $1-1/n$. The weight near each subset of $k$ vertices of $X$ is now more than $k-k/n > k-1$, so again they must be adjacent to at least $k$ vertices of $Y$. Thus, the graph remaining after the removal still satisfies Hall's marriage condition.
What are better bounds for $r(n)$?
 A: Interesting question. If I understand your definitions correctly, this fractional matching in $K_{3,3}$ should show that $r(3) \leq 1/4$. Here, heavy edges have weight $1/2$ in the fractional matching, while light edges have weight $1/4$:

Clearly there is no perfect matching that uses only the heavy edges, so a perfect matching must use edges of weight $1/4$. I haven't thought much more about it, but maybe this example can be generalized to higher $n$ to improve the upper bound.

Trying to keep the constructions in this answer and the computational evidence in the other answer, I think we can generalize this answer to get the upper bound $r(n) \leq 1/\left(\lfloor\frac{n+1}{2}\rfloor\lceil\frac{n+1}{2}\rceil\right)$, which the other answer suggests is probably the right value. When $n$ is even, say $n=2p$, we use the following fractional matching of $K_{n,n}$:

(Here, the boxes represent vertex sets of the given sizes, and the labels on the edges joining the boxes indicate the weights on all edges between those sets.) I believe it should be easy to verify that this a fractional matching, that $1/(p(p+1)) = 1/\left(\lfloor\frac{n+1}{2}\rfloor\lceil\frac{n+1}{2}\rceil\right)$ is the smallest of the nonzero weights, and that there is no perfect matching using only the blue and green edges.
When $n$ is odd, say $n = 2p+1$, I believe a similar construction also works:

I think it should be possible to prove a matching lower bound using LP duality: prior to choosing the values of the $x_{ij}$ variables, the only real choice to make in a vertex cover for the high-weight edges is how many vertices can be used in each part; once that's fixed, all remaining variables are continuous variables, and LP duality should be able to prove that no example with a smaller value of $r$ is possible for the fixed choice of vertex cover. Then it's just a matter of finding a good systematic way to generate dual solutions given the number of vertices of each part in the cover. I haven't thought much about that, but it seems doable.
A: Here is an attempt to formally prove the conjecture of @GregoryJPuleo, namely:
$$r(n) = 1/\left(\lfloor \frac{n+1}{2}\rfloor\lceil\frac{n+1}{2}\rceil\right).$$
We remove from the graph all edges with a weight of less than $r$, and prove that the remaining graph satisfies Hall's marriage condition.
The proof is by contradiction. Let $X_k$ be a subset of $k$ vertices of $X$. Suppose that, after the removal, its set of neighbors is $Y_\ell$ and it contains $\ell\leq k-1$ vertices of $Y$.
Before the removal, the sum of weights near $X_k$ was exactly $k$, and each vertex of $X_k$ had at most $n$ adjacent edges. For each vertex of $X_k$, we had removed at most $n-\ell$ edges to vertices outside $Y_\ell$, and the weight of each such edge is less than $r$; therefore the weight difference between $X_k$ and $Y_\ell$ decreased by less than $k\cdot (n-\ell)\cdot r \leq k\cdot (n-k+1)\cdot r $.
Consider the product $k\cdot (n-k+1)$ as $k$ ranges between $1$ and $n$. It is a product of two integers with a fixed sum $(n+1)$, therefore it is maximized when the two factors are equal up to at most $1$, i.e., when $k = \lfloor \frac{n+1}{2}\rfloor$. Therefore the decrease in weight near $X_k$ is strictly less than
$$\lfloor \frac{n+1}{2}\rfloor \cdot \lceil \frac{n+1}{2}\rceil \cdot r(n) = 1$$
Therefore, the total weight near $X_k$ is strictly more than $k-1$. But this means that $X_k$ must have at least $k$ neighbors in $Y$ - a contradiction.
