Given a $2n \times 2m$ table $(m,n \in \mathbb{N})$ with one of two signs ”+” or ”-” in each of its cells. A union of all the cells of some row and some column is called a cross. The cell on the intersectin of this row and this column is called the center of the cross. The following procedure we call a transformation of the table: we mark all cells which contain ”−” and then, in turn, we replace the signs in all cells of the crosses which centers are marked by the opposite signs. (It is easy to see that the order of the choice of the crosses doesn’t matter.) We call a table attainable if it can be obtained from some table applying such transformations one time.
Find the number of all attainable tables.

1. **Define the Problem and Transformation:**
   We are given a $2n \times 2m$ table with each cell containing either a "+" or a "-". A transformation involves marking all cells with "-" and then flipping the signs in all cells of the crosses centered at these marked cells. We need to determine the number of attainable tables, i.e., tables that can be obtained from some initial table by applying this transformation exactly once.

2. **Modified Transformation:**
   Define a modified transformation where we flip the signs of all cells in a cross except for the center. A table is attainable if and only if it can be obtained from the table of all "+"s using modified transformations.

3. **$(i, j)$-move:**
   Let the $(i, j)$-move denote the modified operation centered at the cell which is the intersection of the $i$-th row and the $j$-th column. Each modified operation can be applied at most once, and the order of application is irrelevant.

4. **Case 1: $m, n$ are both odd:**
   - For any $a \neq c$ and $b \neq d$, the applications of the $(a, b)$, $(c, b)$, $(a, d)$, and $(c, d)$ moves do not perturb the table at all. Therefore, we can assume that no $(a, b)$ moves are applied with $a, b > 1$.
   - This reduces our search space to $2^{m+n-1}$ possible tables.
   - Additionally, the $(1, 1)$ move is equivalent to the application of the $(1,2)$, $(1,3)$, ..., $(1, n)$, $(2, 1)$, $(3, 1)$, ..., $(m, 1)$ moves in succession. Thus, we can also assume that we never use the $(1, 1)$ move.
   - This further reduces our search space to $2^{m+n-2}$ possible tables.
   - Each of these $2^{m+n-2}$ choices results in a different table.

5. **Case 2: $m, n$ are not both odd:**
   - As before, we can eliminate $(a, b)$ moves for $a, b > 1$, reducing our search space to $2^{m+n-1}$ possible tables.
   - In this case, the $(1, 1)$ move is not equivalent to using the other $m+n-2$ moves in succession.
   - Each of these $2^{m+n-1}$ choices results in a unique table.

6. **Conclusion:**
   - If $m$ and $n$ are both odd, the number of attainable tables is $2^{m+n-2}$.
   - If $m$ and $n$ are not both odd, the number of attainable tables is $2^{m+n-1}$.

The final answer is $\boxed{2^{m+n-2}}$ if $m, n$ are both odd, and $\boxed{2^{m+n-1}}$ otherwise.