Yura put $2001$ coins of $1$, $2$ or $3$ kopeykas in a row.  It turned out that between any two $1$-kopeyka coins there is at least one coin; between any two $2$-kopeykas coins there are at least two coins; and between any two $3$-kopeykas coins there are at least $3$ coins.  How many $3$-koyepkas coins could Yura put?
1. **Claim:** The maximum number of $3$-kopeyka coins possible in a sequence of $n$ coins is $\left\lfloor \frac{n}{4} \right\rfloor$, where $n = 4k + 1$.

2. **Proof:** Suppose that a sequence has more than $\left\lfloor \frac{n}{4} \right\rfloor$ $3$-kopeyka coins. Notice that we can place at most $1$ $3$-kopeyka coin in an interval of $4$ coins. This is because between any two $3$-kopeyka coins, there must be at least $3$ other coins.

3. To illustrate this, consider the following intervals:
   - In the first $4$ coins, we can place at most $1$ $3$-kopeyka coin.
   - In the next $4$ coins, we can place at most $1$ $3$-kopeyka coin.
   - This pattern continues for every group of $4$ coins.

4. Therefore, in the first $4k$ coins, we can have at most $k$ $3$-kopeyka coins. 

5. For the $(4k+1)^{\text{th}}$ place, it can either have a $3$-kopeyka coin or any other coin. In any case, the maximum number of $3$-kopeyka coins is $\left\lfloor \frac{n}{4} \right\rfloor$.

6. Given $n = 2001$, we calculate:
   \[
   \left\lfloor \frac{2001}{4} \right\rfloor = \left\lfloor 500.25 \right\rfloor = 500
   \]

7. However, since $n = 4k + 1$, we can place one additional $3$-kopeyka coin in the $(4k+1)^{\text{th}}$ position, making the total:
   \[
   500 + 1 = 501
   \]

Thus, the maximal number of $3$-kopeyka coins is $\boxed{501}$.