Claim: if I construct a bipartite with vertices $R=\{r_1,\cdots,r_{100}\}$ representing rows, $C=\{c_1,\cdots,c_{100}\}$ representing columns. Draw an edge between $r_j$ and $c_k$ if $x_{j,k} \ge l$, then $l$ works if and only if the graph has a matching involving at least 50 rows. Proof: To see this is sufficient, if the graph has a matching $A\to B$ where $A$ is a set of 50 rows, then for each row not in A we add the maximum element in that row (if not already added), and for each column not in B we add the maximum element in that column. This makes sure we pick at most $150$ cells and the sum of elements in any row or column is at least $l$ To see this is necessary we will prove that we can find a matching of 50 in any 150 cells that we pick such that each row and each column has a sum of picked cells at least $l$. Note that if $r_j$ or $c_k$ has exactly one chosen cell, the unique chosen cell on $r_j$ or the unique chosen cell on $c_k$ is at least $l$. Let $S$ be the set of rows that have exactly 1 chosen cell, $T$ be the set of columns that have exactly 1 cell. Let $U$ be the set of chosen cells in both $S$ and $T$; $S_2$ be the set of chosen cells in $T$ but not in $S$, and $T_2$ be the set of chosen cells in $S$ but not in $T$. If $T_2$ cover $k$ columns and $S_2$ cover $m$ rows then there exists a matching of size $|U| + k + m$. Assume FTSOC $|U|+k+m \le 49$ We focus on the $(100-|U|) \times (100-|U|)$ subgrid where the rows and columns containing elements of $U$ are discarded. Consider the the quantity $$X = \# \text{ chosen squares } - \# \text{ rows } - \# \text{ columns } + k + m$$ Note all five things in this expression may change from now on as I remove rows or columns. I will show initially $X\ge 0$, which implies that the number of chosen squares in this subgrid is at least $2(100-|U|)-k-m$. This and the number of squares in $U$ gives a total of $200 - (|U|+k+m)$, so we are done. Suppose a column has at least two cells in $T_2$. Then we remove the column. Then $X$ decreases by at least $2 - 0 - 1 - 1 - 0 + \text{ additional chosen squares not in S or T }$ (this means $k,m$ decreases as well). Furthermore, for each additional square in that column but not in S, we place a +1 in that row. We can remove rows with at least two cells in $S_2$ similarly. After doing this process we want to show that ($X$ + number of +1’s assigned) is at least $a+b-k-m$, where we end up with a $a\times b$ subgrid after removals. For each of the $b-k$ columns that are originally not in $T$, the number of +1’s plus number of remaining chosen squares is at least 2. So $X$ + number of +1’s assigned is at least $2(b-k)$. Similarly it is at least $2(a-m)$, so we are done. The answer is $\frac{17}{1900}$ Construction: let $x_{j,k}=0$ if $1\le j\le 25, 1\le k\le 24$ $x_{j,k}=\frac{1}{75}$ if $26\le j\le 100, 1\le k\le 24$ $x_{j,k}=\frac{1}{76}$ if $1\le j\le 25, 25\le k\le 100$ $x_{j,k}=\frac{17}{1900}$ if $26\le j\le 100, 25\le k\le 100$ We can see that for any $\lambda > \frac{17}{1900}$. Proof of Optimality: Consider a bipartite graph with vertices $\{r_1,\cdots,r_{100}\}$ representing rows, $\{c_1,\cdots,c_{100}\}$ representing columns. Draw an edge between $r_j$ and $c_k$ if $x_{j,k} \ge \frac{17}{1900}$. It suffices to prove there exists a matching of size at least 50. Let S be a set of rows such that $|N(S)|-|S|$ is minimized. I claim $|N(S)|-|S| \ge -50$ Proof: The set of cells in $S \cap N(S)$ have sum greater than $|N(S)|$ by algebra. With this in mind, note that we can biject $R\setminus S$ to $C \setminus N(S)$ because if not, hall condition is violated, so for some $T \subset R \setminus S$, there are at most $|T|-1$ columna in $C\backslash N(S)$ that has a neighbour in $T$, then $|N(S\sqcup T)|-|S \sqcup T| =( |N(S)| - |S|) + (|N(T)\setminus N(S)| - |T|) < ( |N(S)| - |S|)$ contradicting the minimality of |N(S)|-|S|. We can also construct an injection from $N(S)$ to $S$ because otherwise otherwise, say some $U\subset N(S)$ has $|N(U)|<|U|$, then $N(S \backslash N(U)) \subset N(S)\setminus U$ and we are done by minimality. This allows us to construct a matching of size at least $|N(S)|+|R\setminus S| = |R| - (|S|-|N(S)|) =50$