Problem 1340

1. Initial Setup:
- We start with 100 piles of 100 stones each.
- We need to find the largest $n \leq 10,000$ such that after removing any $n$ stones, there will still be 50 piles numbered from 1 to 50, where the $i$-th pile has at least $i$ stones.

2. Part 1:
- We can remove 5100 stones in such a way that the condition is not satisfied.
- Specifically, if we remove 51 stones from each pile, we will have removed $51 \times 100 = 5100$ stones.
- After this removal, each pile will have $100 - 51 = 49$ stones.
- Since the 50th pile needs at least 50 stones, the condition is not satisfied.

3. Part 2:
- We need to show that if we remove 5099 stones, the condition is always satisfied.
- Suppose we remove $k_i$ stones from the $i$-th pile, with $k_1 \geq k_2 \geq \ldots \geq k_{100}$.
- We claim that $k_i \leq 150 - i$ for all $i \in \{51, 52, \ldots, 100\}$.

4. Proof of the Claim:
- Assume there exists an $i \in \{51, \ldots, 100\}$ such that $k_i \geq 151 - i$.
- Then, we have:
$$5099 = k_1 + k_2 + \ldots + k_{100} \geq k_1 + k_2 + \ldots + k_i \geq i \cdot k_i \geq i \cdot (151 - i)$$
- We need to show that $i \cdot (151 - i) \geq 5100$ leads to a contradiction.
- Consider the function $f(i) = i \cdot (151 - i)$:
$$f(i) = 151i - i^2$$
- This is a quadratic function opening downwards with a maximum at $i = \frac{151}{2} \approx 75.5$.
- Evaluating $f(i)$ at the boundaries $i = 51$ and $i = 100$:
$$f(51) = 51 \cdot (151 - 51) = 51 \cdot 100 = 5100$$
$$f(100) = 100 \cdot (151 - 100) = 100 \cdot 51 = 5100$$
- Since $f(i) \leq 5100$ for $51 \leq i \leq 100$, the assumption $k_i \geq 151 - i$ leads to $5099 \geq 5100$, which is a contradiction.
- Therefore, $k_i \leq 150 - i$ for all $i \in \{51, 52, \ldots, 100\}$.

5. Verification:
- After removing $k_i$ stones, each pile $i$ has $100 - k_i$ stones left.
- For $i \in \{51, 52, \ldots, 100\}$:
$$100 - k_i \geq 100 - (150 - i) = i - 50$$
- Thus, the 51st pile has at least 1 stone, the 52nd pile has at least 2 stones, ..., and the 100th pile has at least 50 stones.

Conclusion:
$$\boxed{5099}$$