Let $1,7,19,\ldots$ be the sequence of numbers such that for all integers $n\ge 1$, the average of the first $n$ terms is equal to the $n$th perfect square. Compute the last three digits of the $2021$st term in the sequence.

[i]Proposed by Nathan Xiong[/i]
1. Let \( x_n \) be the \( n \)-th term and \( s_n \) be the sum of the first \( n \) terms for all \( n \in \mathbb{Z}_{+} \). Since the average of the first \( n \) terms is \( n^2 \), we have:
   \[
   \frac{s_n}{n} = n^2 \implies s_n = n^3
   \]
   This implies that the sum of the first \( n \) terms is \( n^3 \).

2. To find the \( n \)-th term \( x_n \), we use the relationship:
   \[
   x_n = s_n - s_{n-1}
   \]
   where \( s_n = n^3 \) and \( s_{n-1} = (n-1)^3 \). Therefore:
   \[
   x_n = n^3 - (n-1)^3
   \]

3. We expand \( n^3 - (n-1)^3 \) using the binomial theorem:
   \[
   n^3 - (n-1)^3 = n^3 - (n^3 - 3n^2 + 3n - 1) = 3n^2 - 3n + 1
   \]

4. We need to compute the last three digits of the \( 2021 \)-st term in the sequence, \( x_{2021} \):
   \[
   x_{2021} = 2021^3 - 2020^3
   \]

5. To simplify the calculation modulo 1000, we use:
   \[
   2021 \equiv 21 \pmod{1000} \quad \text{and} \quad 2020 \equiv 20 \pmod{1000}
   \]
   Therefore:
   \[
   x_{2021} \equiv 21^3 - 20^3 \pmod{1000}
   \]

6. Calculate \( 21^3 \) and \( 20^3 \):
   \[
   21^3 = 21 \times 21 \times 21 = 441 \times 21 = 9261
   \]
   \[
   20^3 = 20 \times 20 \times 20 = 400 \times 20 = 8000
   \]

7. Subtract the results modulo 1000:
   \[
   21^3 - 20^3 \equiv 9261 - 8000 \equiv 1261 - 1000 \equiv 261 \pmod{1000}
   \]

8. Therefore, the last three digits of the \( 2021 \)-st term are:
   \[
   \boxed{261}
   \]