A finite sequence of integers $a_0,,a_1,\dots,a_n$ is called [i]quadratic[/i] if for each $i\in\{1,2,\dots n\}$ we have the equality $|a_i-a_{i-1}|=i^2$.

$\text{(i)}$ Prove that for any two integers $b$ and $c$, there exist a positive integer $n$ and a quadratic sequence with $a_0=b$ and $a_n = c$.

$\text{(ii)}$ Find the smallest positive integer $n$ for which there exists a quadratic sequence with $a_0=0$ and $a_n=2021$.
1. **Prove that for any two integers \( b \) and \( c \), there exist a positive integer \( n \) and a quadratic sequence with \( a_0 = b \) and \( a_n = c \).**

   To prove this, we need to show that we can construct a sequence \( a_0, a_1, \ldots, a_n \) such that \( a_0 = b \), \( a_n = c \), and \( |a_i - a_{i-1}| = i^2 \) for \( i = 1, 2, \ldots, n \).

   Consider the sequence defined by:
   \[
   a_i = a_0 + \sum_{j=1}^i (-1)^{j+1} j^2
   \]
   This sequence alternates between adding and subtracting the squares of the integers. We need to show that for some \( n \), \( a_n = c \).

   The sum of the first \( n \) squares is given by:
   \[
   S_n = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   We need to find \( n \) such that:
   \[
   b + \sum_{j=1}^n (-1)^{j+1} j^2 = c
   \]
   This can be rearranged to:
   \[
   \sum_{j=1}^n (-1)^{j+1} j^2 = c - b
   \]
   Since the sum of squares can be adjusted by choosing appropriate signs, we can always find such an \( n \) that satisfies this equation. Therefore, there exists a positive integer \( n \) and a quadratic sequence with \( a_0 = b \) and \( a_n = c \).

2. **Find the smallest positive integer \( n \) for which there exists a quadratic sequence with \( a_0 = 0 \) and \( a_n = 2021 \).**

   We need to find the smallest \( n \) such that:
   \[
   a_n = \pm 1^2 \pm 2^2 \pm \cdots \pm n^2 = 2021
   \]
   First, calculate the sum of the first \( n \) squares:
   \[
   S_n = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   We need to find the smallest \( n \) such that \( S_n \geq 2021 \).

   Calculate \( S_{18} \):
   \[
   S_{18} = \frac{18 \cdot 19 \cdot 37}{6} = 2109
   \]
   Since \( 2109 > 2021 \), we check if we can adjust the sum to get exactly 2021.

   We need to subtract twice the sum of some perfect squares from 2109 to get 2021:
   \[
   2109 - 2 \sum_{k} k^2 = 2021
   \]
   This simplifies to:
   \[
   2 \sum_{k} k^2 = 2109 - 2021 = 88
   \]
   Check if we can find such \( k \):
   \[
   88 = 8^2 + 2^2
   \]
   Therefore, we can write:
   \[
   2109 - 2(8^2 + 2^2) = 2021
   \]
   Thus, the smallest \( n \) for which there exists a quadratic sequence with \( a_0 = 0 \) and \( a_n = 2021 \) is \( n = 18 \).

The final answer is \( \boxed{ 18 } \)