Solution:

The competitor who takes the largest slice is the one who serves himself 100th.
For every $k=1,2, \ldots, 10000$, let us denote by $F_{k}$ the slice belonging to the $k$-th competitor, and by $T_{k}$ the amount of cake remaining after the first $k$ competitors have served themselves. Let us set by convention $T_{0}=1$. We must determine the value of $k$ for which $F_{k}$ is maximum. The statement of the problem tells us that
$$F_{k}=\frac{k}{n} \cdot T_{k-1} \quad \text{and} \quad T_{k}=\frac{n-k}{n} \cdot T_{k-1} .$$
From these two equations we derive that
$$F_{k+1}=\frac{k+1}{n} \cdot T_{k}=\frac{k+1}{n} \cdot \frac{n-k}{n} \cdot T_{k-1},$$
and hence
$$\frac{F_{k+1}}{F_{k}}=\frac{k+1}{n} \cdot \frac{n-k}{n} \cdot \frac{n}{k}=\frac{(k+1)(n-k)}{n k}=1-\frac{k(k+1)-n}{k n} .$$
This tells us in particular that $F_{k+1}$ is greater than $F_{k}$ if and only if
$$k(k+1)<n .$$
Taking into account that $n=10000$, this happens if and only if $k \leq 99$ (it suffices to observe that $k(k+1)$ is an increasing function of the positive integer $k$ and the inequalities $99 \cdot 100<10000 \mathrm{and}$ $100 \cdot 101>10000$ hold).
It follows that the value of $F_{k}$ initially increases until it reaches the maximum when $k=100$, and then decreases from there on until the end of the competitors.