Each cell of a $29 \times 29$ table contains one of the integers $1, 2, 3, \ldots , 29$, and each of these integers appears $29$ times. The sum of all the numbers above the main diagonal is equal to three times the sum of all the numbers below this diagonal. Determine the number in the central cell of the table.
1. **Understanding the Problem:**
   - We have a \(29 \times 29\) table.
   - Each integer from \(1\) to \(29\) appears exactly \(29\) times.
   - The sum of all numbers above the main diagonal is three times the sum of all numbers below the main diagonal.
   - We need to determine the number in the central cell of the table.

2. **Analyzing the Diagonal:**
   - The main diagonal consists of \(29\) cells.
   - The cells above the main diagonal are \(\frac{29 \times 28}{2} = 406\) cells.
   - The cells below the main diagonal are also \(406\) cells.

3. **Sum of Numbers:**
   - The sum of all numbers in the table is:
     \[
     29 \times \left(1 + 2 + \ldots + 29\right) = 29 \times \frac{29 \times 30}{2} = 29 \times 435 = 12615
     \]

4. **Sum of Numbers Above and Below the Diagonal:**
   - Let \(S_{\text{below}}\) be the sum of the numbers below the diagonal.
   - Let \(S_{\text{above}}\) be the sum of the numbers above the diagonal.
   - Given \(S_{\text{above}} = 3 \times S_{\text{below}}\).

5. **Sum of Numbers on the Diagonal:**
   - Let \(S_{\text{diag}}\) be the sum of the numbers on the diagonal.
   - We have:
     \[
     S_{\text{diag}} + S_{\text{above}} + S_{\text{below}} = 12615
     \]
   - Substituting \(S_{\text{above}} = 3 \times S_{\text{below}}\):
     \[
     S_{\text{diag}} + 3 \times S_{\text{below}} + S_{\text{below}} = 12615
     \]
     \[
     S_{\text{diag}} + 4 \times S_{\text{below}} = 12615
     \]

6. **Finding \(S_{\text{below}}\) and \(S_{\text{diag}}\):**
   - The sum of the numbers from \(1\) to \(14\) is:
     \[
     1 + 2 + \ldots + 14 = \frac{14 \times 15}{2} = 105
     \]
   - The sum of the numbers from \(16\) to \(29\) is:
     \[
     16 + 17 + \ldots + 29 = \frac{29 \times 30}{2} - \frac{15 \times 16}{2} = 435 - 120 = 315
     \]
   - Since there are \(29\) instances of each number, the total sum below the diagonal is:
     \[
     S_{\text{below}} = 29 \times 105 = 3045
     \]
   - The total sum above the diagonal is:
     \[
     S_{\text{above}} = 3 \times 3045 = 9135
     \]
   - Therefore, the sum on the diagonal is:
     \[
     S_{\text{diag}} = 12615 - 9135 - 3045 = 435
     \]

7. **Finding the Central Cell:**
   - The central cell is the 15th cell on the diagonal (since \(29\) is odd, the central cell is the middle one).
   - Since the sum of the numbers on the diagonal is \(435\) and there are \(29\) cells, the average value of each cell on the diagonal is:
     \[
     \frac{435}{29} = 15
     \]
   - Therefore, the number in the central cell is \(15\).

The final answer is \(\boxed{15}\).