On the table, there're $1000$ cards arranged on a circle. On each card, a positive integer was written so that all $1000$ numbers are distinct. First, Vasya selects one of the card, remove it from the circle, and do the following operation: If on the last card taken out was written positive integer $k$, count the $k^{th}$ clockwise card not removed, from that position, then remove it and repeat the operation. This continues until only one card left on the table. Is it possible that, initially, there's a card $A$ such that, no matter what other card Vasya selects as first card, the one that left is always card $A$?
1. **Initial Setup**: Consider a circle with 1000 cards, each labeled with a distinct positive integer. We need to determine if there exists a card \( A \) such that no matter which card Vasya starts with, the last remaining card is always \( A \).

2. **Simplified Case**: Let's first simplify the problem by temporarily ignoring the distinctness condition. Assume we have two adjacent cards \( A \) and \( B \), where \( A \) is immediately clockwise from \( B \). Assign the number 1 to all cards except \( B \), which is assigned the number 2.

3. **Operation Analysis**:
   - If Vasya removes any card other than \( A \) first, he will continue removing cards in a clockwise manner.
   - When he reaches \( B \), the number 2 on \( B \) will cause him to skip \( A \) and continue removing the next card.
   - This skipping ensures that \( A \) is never removed until all other cards are removed.

4. **Ensuring Distinct Numbers**: To satisfy the condition that all numbers on the cards are distinct, we can add distinct multiples of \( 1000! \) to each card. This ensures that:
   - The relative order of removal remains unchanged because adding multiples of \( 1000! \) does not affect the counting process modulo 1000.
   - Each card will have a unique number since \( 1000! \) is a very large number, and adding distinct multiples of it will ensure distinctness.

5. **Conclusion**: By assigning numbers in this manner, we ensure that card \( A \) will always be the last remaining card, regardless of which card Vasya starts with.

\[
\boxed{\text{True}}
\]