In a mathematical competition $n=10\,000$ contestants participate.
During the final party, in sequence, the first one takes $1/n$ of the cake, the second one takes $2/n$ of the remaining cake, the third one takes $3/n$ of the cake that remains after the first and the second contestant, and so on until the last one, who takes all of the remaining cake.
Determine which competitor takes the largest piece of cake.
1. **Initial Setup:**
   We start with \( n = 10,000 \) contestants. The first contestant takes \( \frac{1}{n} \) of the cake, the second takes \( \frac{2}{n} \) of the remaining cake, and so on.

2. **General Formula:**
   Let \( a_k \) be the amount of cake remaining after the \( k \)-th contestant has taken their share. Initially, \( a_0 = 1 \) (the whole cake).

   The \( k \)-th contestant takes \( \frac{k}{n} \) of the remaining cake \( a_{k-1} \):
   \[
   \text{Amount taken by the } k\text{-th contestant} = \frac{k}{n} a_{k-1}
   \]
   The remaining cake after the \( k \)-th contestant is:
   \[
   a_k = a_{k-1} - \frac{k}{n} a_{k-1} = a_{k-1} \left(1 - \frac{k}{n}\right)
   \]

3. **Amount Taken by Each Contestant:**
   The amount taken by the \( k \)-th contestant is:
   \[
   \frac{k}{n} a_{k-1}
   \]
   The amount taken by the \( (k+1) \)-th contestant is:
   \[
   \frac{k+1}{n} a_k = \frac{k+1}{n} a_{k-1} \left(1 - \frac{k}{n}\right)
   \]

4. **Comparing Pieces:**
   To determine which contestant takes the largest piece, we compare the pieces taken by consecutive contestants. We need to find \( k \) such that:
   \[
   \frac{k}{n} a_{k-1} \geq \frac{k+1}{n} a_k
   \]
   Substituting \( a_k = a_{k-1} \left(1 - \frac{k}{n}\right) \):
   \[
   \frac{k}{n} a_{k-1} \geq \frac{k+1}{n} a_{k-1} \left(1 - \frac{k}{n}\right)
   \]
   Simplifying, we get:
   \[
   k \geq (k+1) \left(1 - \frac{k}{n}\right)
   \]
   \[
   k \geq (k+1) - \frac{(k+1)k}{n}
   \]
   \[
   k \geq k + 1 - \frac{k^2 + k}{n}
   \]
   \[
   0 \geq 1 - \frac{k^2 + k}{n}
   \]
   \[
   \frac{k^2 + k}{n} \geq 1
   \]
   \[
   k^2 + k \geq n
   \]
   Substituting \( n = 10,000 \):
   \[
   k^2 + k \geq 10,000
   \]

5. **Solving the Quadratic Inequality:**
   We solve the quadratic inequality \( k^2 + k - 10,000 \geq 0 \):
   \[
   k = \frac{-1 \pm \sqrt{1 + 4 \cdot 10,000}}{2} = \frac{-1 \pm \sqrt{40,001}}{2}
   \]
   Approximating \( \sqrt{40,001} \approx 200 \):
   \[
   k \approx \frac{-1 + 200}{2} = 99.5
   \]
   Since \( k \) must be an integer, we consider \( k = 99 \) and \( k = 100 \).

6. **Conclusion:**
   For \( k = 99 \):
   \[
   99^2 + 99 = 9,801 + 99 = 9,900 < 10,000
   \]
   For \( k = 100 \):
   \[
   100^2 + 100 = 10,000 + 100 = 10,100 > 10,000
   \]
   Therefore, the largest piece is taken by the \( 100 \)-th contestant.

The final answer is \( \boxed{100} \)