Solution:

We start with the following known consequence of Hall's lemma.

Lemma. Let $G=(U \sqcup V, E)$ be a bipartite multigraph with parts $U$ and $V$, both of size $n$. Assume that each vertex has degree $k$; then the edges can be partitioned into $k$ perfect matchings.

Proof. Induction on $k$; the base case $k=1$ is trivial. To perform the step, it suffices to find one perfect matching in the graph: removing the edges of that matching, we obtain a graph with all degrees equal to $k-1$.

The existence of such matching is guaranteed by Hall's lemma. Indeed, let $U'$ be any subset of $U$, and let $V'$ be the set of vertices adjacent to $U'$. Put $u=|U'|$ and $v=|V'|$. The total degree of vertices in $U'$ is $k u$, so the total degree of vertices in $V'$ is at least $k u$; hence $k u \leq k v$ and therefore $u \leq v$, which establishes the conditions of Hall's lemma.

The following claim is the principal step in this solution.

Claim. In any good table, one can decrease some numbers so that the table becomes balanced.

Proof. Say that a cell in a good table is blocked if it is contained in a vanishing rook set (so, decreasing the number in the cell would break goodness of the table). First, we show that in any good table one can decrease several numbers so that the table remains good, and all its cells become blocked.

Consider any cell $c$; let $\epsilon$ be the minimal sum in a rook set containing that cell. Decrease the number in $c$ by $\epsilon$; the obtained table is still good, but now $c$ is blocked. Apply such operation to all cells in the table consecutively; we arrive at a good table all whose cells are blocked. We claim that, in fact, this table is balanced.

In the sequel, we use the following correspondence. Let $R$ and $C$ be the sets of rows and columns of the table, respectively. Then each cell corresponds to a pair of the row and the column it is situated in; this pair may be regarded as an edge of a bipartite (multi)graph with parts $R$ and $C$. This way, any rook set corresponds to a perfect matching between those parts.

Arguing indirectly, assume that there is a non-vanishing rook set $S=\{s_1, s_2, \ldots, s_n\}$. Each cell $s_i$ is contained in some vanishing rook set $V_i$. Now construct a bipartite multigraph $G=(R \sqcup C, E)$, introducing, for each set $V_i$, $n$ edges corresponding to its cells (thus, $G$ contains $n^2$ edges some of which may be parallel).

Mark each edge with the number in the corresponding cell. Since the sets $V_i$ are all vanishing, the sum of all $n^2$ marks is zero.

Now, remove $n$ edges corresponding to the cells of $S$, to obtain a graph $G'$. Since the sum of numbers in the cells of $S$ is positive, the sum of the marks in $G'$ is negative. On the other hand, the degree of every vertex in $G'$ is $n-1$, so by the Lemma its edges can be partitioned into $n-1$ perfect matchings. At least one of the obtained matchings has negative sum of marks; so this matching corresponds to a rook set with a negative sum. This is impossible in a good table; this contradiction finishes the proof.

Back to the problem, let $T$ be Elwyn's table. Applying the Claim, decrease some numbers in it to get a balanced table $B$. By Proposition 2, Elwyn can perform some moves on table $B$ so as to get a table filled with zeros. Applying the same moves to $T$, Elwyn gets a table where all numbers are nonnegative, as required.

Solution:

Say that the badness of a table is the sum of absolute values of all its negative entries. In Step 1, we will show that, whenever the badness of a good table is nonzero, Elwyn can make some moves decreasing the badness. In a (technical) Step 2, we will show that this claim yields the required result.

Step 1. Let $r$ be a row containing some negative number. Mark all cells in row $r$ containing negative numbers, and mark all cells in other rows containing nonpositive numbers. Then there is no rook set consisting of marked cells, since that set would not be nonnegative.

By K\"onig's theorem (which is equivalent to Hall's lemma), for some $a$ and $b$ with $a+b<n$, one can choose $a$ rows and $b$ columns such that their union contains all marked cells; fix such a choice. Number the rows from top to bottom, and the columns from left to right. We distinguish two cases.

Case 1: Row $r$ is among the $a$ chosen rows.

Permute the rows and columns so that the top $a$ rows and the right $b$ columns are chosen. Next, if row $r$ contains a negative number in some of the $a$ leftmost entries, swap the column containing that entry with the $(n-b)$th one (recall that $n-b>a$). As a result, there exists $x>a$ such that the $x$th left entry in row $r$ is negative (while the chosen columns are still the $b$ rightmost ones).

Now, rectangle $P$ formed by the bottom $n-a$ rows and the left $a$ columns contains only positive numbers, as it contains no marked cells, as well as no cells from row $r$. Let $m$ be the minimal number in that rectangle.

Let Elwyn add $m$ to all numbers in the first $a$ rows, and subtract $m$ from all numbers in the first $a$ columns. All numbers which decrease after this operation are situated in $P$, so there appear no new cell containing a negative number, and no negative number decreases. Moreover, by our choice, at least one negative number (situated in row $r$ and column $x$) increases. Thus, the badness decreases, as desired.

Case 2: Row $r$ is not among the $a$ chosen rows.

Add row $r$ to the $a$ chosen rows, and increase $a$ by 1. Notice that the negative numbers in row $r$ are covered by the $b$ chosen columns. As in the previous case, we permute the rows and columns so that the top $a$ rows and the right $b$ columns are chosen. All negative numbers in row $r$ automatically come to the right $b$ columns. Now the above argument applies verbatim.

Step 2. We show that among the tables which Elwyn can obtain (call such tables reachable), there exists a table with the smallest badness. Applying the argument in Step 1 to that table, we get that its badness is zero, which proves the claim of the problem.

Notice that the effect of any sequence of Elwyn's moves has the form described in Proposition 1. Moreover, subtraction of some number $\epsilon$ from all the $a_i$ and the $b_i$ provides no effect on the result. Hence, we may assume that the sums of the $a_i$ and of the $b_i$ are both zero.

Let $t_{ij}$ denote the $(i, j)$th entry of the initial table $T$. For any two sequences $\mathbf{a}=(a_1, \ldots, a_n)$ and $\mathbf{b}=(b_1, \ldots, b_n)$ both summing up to zero, denote by $T(\mathbf{a}, \mathbf{b})$ the table obtained from $T$ by adding $a_i$ to all numbers in the $i$th row, and subtracting $b_j$ from all numbers in the $j$th column, for all $i, j=1,2, \ldots, n$; in particular, $T=T(\mathbf{0}, \mathbf{0})$, where $\mathbf{0}=(0,0, \ldots, 0)$. Let $f(\mathbf{a}, \mathbf{b})$ denote the badness of $T(\mathbf{a}, \mathbf{b})$. Clearly, function $f$ is continuous. Now we intend to bound the set of values that make sense to put in sequences $\mathbf{a}$ and $\mathbf{b}$.

Let $m$ be the maximal number in $T$. Take any $\mathbf{a}$ and $\mathbf{b}$ summing up to zero, such that some $a_i$ is smaller than $-M=-(m+b)$. Then there exists an index $j$ with $b_j \geq 0$; hence the entry $(i, j)$ in $T(\mathbf{a}, \mathbf{b})$ is $t_{ij}+a_i-b_j<m-M+0=-b$, so $f(\mathbf{a}, \mathbf{b})>b=f(\mathbf{0}, \mathbf{0})$.

So, all pairs of sequences $\mathbf{a}$ and $\mathbf{b}$ satisfying $f(\mathbf{a}, \mathbf{b}) \leq b$ should also satisfy $a_i \geq -M$ and $b_j \geq -M$, and hence $a_i \leq n M$ and $b_j \leq n M$ as well (since each of the sequences sums up to zero). Thus, in order to minimize $f(\mathbf{a}, \mathbf{b})$, it suffices to consider only those $\mathbf{a}$ and $\mathbf{b}$ whose entries lie in $[-M, n M]$. Those values form a compact set, so the continuous function $f$ attains the smallest value on that set.

Solution:

We implement some tools from multi-dimensional convex geometry.

Each table can be regarded as a point in $\mathbb{R}^{n \times n}$. The set $G$ of good tables is a convex cone determined by $n!$ non-strict inequalities (claiming that the rook sets are nonnegative). Thus this cone is closed.

The set $T$ of tables which can be transformed, by a sequence of Elwyn's moves, into a table with nonnegative entries, is also a convex cone. This cone is the Minkowski sum of the (closed) cone $N$ of all tables with nonnegative entries and the linear subspace $V$ of all tables Elwyn can add by a sequence of moves. Such sum is always closed (a pedestrian version of such argument is presented in Step 2 of Solution 2).

It is easy to see that $T \subseteq G$; we need to show that $T=G$. Arguing indirectly, assume that there is some table $t \in G \setminus T$. Then there exists a linear function $f$ separating $t$ and $T$, that is $-f$ takes nonnegative values on $T$ but a negative value on $t$.

This function $f$ has the following form: Let $x \in \mathbb{R}^{n \times n}$ be a table, and denote by $x_{ij}$ its $(i, j)$th entry. Then
$$f(x)=\sum_{i, j=1}^{n} f_{ij} x_{ij}$$
where $f_{ij}$ are some real constants. Form a table $F$ whose $(i, j)$th entry is $f_{ij}$.

Since $f(x) \geq 0$ for all tables in $N$ having only one nonnegative entry, we have $f_{ij} \geq 0$ for all $i$ and $j$. Moreover, $f$ must vanish on all tables in the subspace $V$, in particular - on each table having 1 in some row, -1 in some column, and 0 elsewhere (the intersection of the row and the column also contains 0). This means that the sum of numbers in any row in $F$ is equal to the sum of the numbers in any its column.

Now it remains to show that $F$ is the sum of several rook tables which contain some nonnegative number $p$ at the cells of some rook set, while all other entries are zero; this will yield $f(t) \geq 0$ which is not the case. In other words, it suffices to prove that one can subtract from $F$ several rook tables to make it vanish. This can be done by means of Hall's lemma again: if the table is still nonzero, it contains $n$ positive entries forming a rook set, and one may make one of them vanish, keeping the other entries nonnegative, by subtracting a rook table.