Every bacterium has a horizontal body with natural length and some nonnegative number of vertical feet, each with nonnegative (!) natural length, that lie below its body. In how many ways can these bacteria fill an $m\times n$ table such that no two of  them overlap?

[i]proposed by Mahyar Sefidgaran[/i]
To solve the problem of determining the number of ways bacteria can fill an \( m \times n \) table such that no two of them overlap, we need to consider the placement of the bacteria's bodies and their feet. Here is a detailed step-by-step solution:

1. **Understanding the Bacteria Structure**:
   - Each bacterium has a horizontal body of natural length.
   - Each bacterium has a nonnegative number of vertical feet, each with nonnegative natural length, that lie below its body.

2. **Placing the Bacteria's Bodies**:
   - The bacteria's bodies must be placed in the \( m \times n \) table such that no two bodies overlap.
   - We need to count the number of ways to place these bodies in the table.

3. **Recurrence Relation**:
   - The problem can be approached using a recurrence relation. The placement of the bacteria's bodies can be thought of as a combinatorial problem involving Fibonacci numbers.
   - Let \( f_k \) denote the \( k \)-th Fibonacci number, where \( f_0 = 0 \) and \( f_1 = 1 \).

4. **Counting the Arrangements**:
   - The number of ways to place the bacteria's bodies in the \( n \)-length row is given by \( 2^{n-1} \). This is because each position in the row can either be occupied by a part of a bacterium's body or be empty.
   - For each row, the number of ways to place the bacteria's bodies is \( 2^{n-1} \).

5. **Considering the Feet**:
   - The feet of the bacteria can be placed in the remaining space below the bodies.
   - The number of ways to place the feet in the \( m \)-length column is given by \( (f_{2n+1})^{m-1} \). This is because the feet can be placed in a way that corresponds to the Fibonacci sequence.

6. **Combining the Results**:
   - The total number of ways to fill the \( m \times n \) table with bacteria such that no two of them overlap is given by the product of the number of ways to place the bodies and the number of ways to place the feet.

\[
\text{Total number of ways} = 2^{n-1} \cdot (f_{2n+1})^{m-1}
\]

The final answer is \(\boxed{2^{n-1} \cdot (f_{2n+1})^{m-1}}\).