There are 20 people at a party. Each person holds some number of coins. Every minute, each person who has at least 19 coins simultaneously gives one coin to every other person at the party. (So, it is possible that $A$ gives $B$ a coin and $B$ gives $A$ a coin at the same time.) Suppose that this process continues indefinitely. That is, for any positive integer $n$, there exists a person who will give away coins during the $n$th minute. What is the smallest number of coins that could be at the party?

[i]Proposed by Ray Li[/i]
1. **Initial Setup and Distribution**:
   - We start with 20 people at a party, each holding some number of coins.
   - Every minute, each person who has at least 19 coins gives one coin to every other person at the party.
   - The process continues indefinitely, meaning for any positive integer \( n \), there exists a person who will give away coins during the \( n \)-th minute.

2. **Finding the Minimum Number of Coins**:
   - We need to determine the smallest number of coins that could be at the party.
   - Consider the initial distribution of coins as \( 19, 18, \ldots, 2, 1, 0 \). This distribution sums up to:
     \[
     \sum_{i=0}^{19} i = \frac{19 \cdot 20}{2} = 190
     \]
   - This distribution ensures that every minute, at least one person will have 19 or more coins to give away, thus satisfying the condition that the process continues indefinitely.

3. **Lower Bound Proof**:
   - To show that 190 is the minimum number of coins required, we need to prove that any distribution with fewer than 190 coins cannot sustain the process indefinitely.
   - Split the sequence into turns where only one person donates. If many people want to donate simultaneously, we can arbitrarily order the donations. The final position will be the same as if all donated at the same time.

4. **Analyzing Donations**:
   - Let \( a_p \) denote the number of times person \( p \) donates in the first 19 turns.
   - Let \( x_p \) be the last turn person \( p \) donates on.
   - If person \( p \) started with \( c_p \) coins, then:
     \[
     c_p + x_p - a_p \ge 19a_p
     \]
     This is because \( x_p - a_p \) is the number of coins they gain until their last turn, and overall they ended up giving away \( 19a_p \) coins.
   - Therefore:
     \[
     c_p \ge 20a_p - x_p
     \]

5. **Summing Up**:
   - The total number of coins \( C \) is:
     \[
     C = \sum_{p=1}^{20} c_p \ge \sum_{p=1}^{20} (20a_p - x_p) = 20 \sum_{p=1}^{20} a_p - \sum_{p=1}^{20} x_p
     \]
   - Since each person can donate at most once per turn in the first 19 turns:
     \[
     \sum_{p=1}^{20} a_p \le 19
     \]
   - The sum of the last turns \( x_p \) for each person is:
     \[
     \sum_{p=1}^{20} x_p \le 1 + 2 + \cdots + 19 = \frac{19 \cdot 20}{2} = 190
     \]
   - Therefore:
     \[
     C \ge 20 \cdot 19 - 190 = 380 - 190 = 190
     \]

Conclusion:
- The smallest number of coins that could be at the party is \( \boxed{190} \).