A finite sequence of integers $a_{0}, a_{1}, \cdots, a_{n}$ is called quadratic if for each $i \in \{1,2,\cdots,n \}$ we have the equality $\vert a_{i}-a_{i-1} \vert = i^2$. [list=a] [*] Prove that for any two integers $b$ and $c$, there exists a natural number $n$ and a quadratic sequence with $a_{0}=b$ and $a_{n}=c$. [*] Find the smallest natural number $n$ for which there exists a quadratic sequence with $a_{0}=0$ and $a_{n}=1996$. [/list]
### Part (a)
1. **Restate the problem**: We need to prove that for any two integers \( b \) and \( c \), there exists a natural number \( n \) and a quadratic sequence \( a_0, a_1, \ldots, a_n \) such that \( a_0 = b \) and \( a_n = c \).

2. **Define the sequence**: A sequence is quadratic if for each \( i \in \{1, 2, \ldots, n\} \), we have \( |a_i - a_{i-1}| = i^2 \).

3. **Express the difference**: Let \( d_i = a_i - a_{i-1} \). Then \( |d_i| = i^2 \).

4. **Sum of differences**: The total change from \( a_0 \) to \( a_n \) is:
   \[
   a_n - a_0 = \sum_{i=1}^n d_i
   \]
   We need to show that \( \sum_{i=1}^n d_i = c - b \).

5. **Possible values of \( d_i \)**: Since \( |d_i| = i^2 \), \( d_i \) can be either \( i^2 \) or \( -i^2 \).

6. **Constructing the sum**: We need to show that any integer \( m \) can be written as:
   \[
   m = \pm 1^2 \pm 2^2 \pm 3^2 \pm \ldots \pm n^2
   \]
   This is equivalent to showing that any integer can be expressed as a sum of signed squares of the first \( n \) natural numbers.

7. **Using induction**: We can use induction to show that any integer can be written in this form. For the base case, \( n = 1 \), we have \( \pm 1^2 = \pm 1 \), which covers all integers of the form \( \pm 1 \).

8. **Inductive step**: Assume that any integer \( m \) can be written as \( \pm 1^2 \pm 2^2 \pm \ldots \pm k^2 \). For \( n = k+1 \), we need to show that any integer \( m \) can be written as \( \pm 1^2 \pm 2^2 \pm \ldots \pm k^2 \pm (k+1)^2 \). This can be done by adjusting the sign of \( (k+1)^2 \) to achieve the desired sum.

9. **Conclusion**: By induction, any integer \( m \) can be written as \( \pm 1^2 \pm 2^2 \pm \ldots \pm n^2 \) for some \( n \). Therefore, for any integers \( b \) and \( c \), there exists a natural number \( n \) and a quadratic sequence with \( a_0 = b \) and \( a_n = c \).

\(\blacksquare\)

### Part (b)
1. **Restate the problem**: Find the smallest natural number \( n \) for which there exists a quadratic sequence with \( a_0 = 0 \) and \( a_n = 1996 \).

2. **Sum of squares**: We need to find the smallest \( n \) such that:
   \[
   \sum_{i=1}^n i^2 \geq 1996
   \]

3. **Calculate the sum of squares**: The sum of the first \( n \) squares is given by:
   \[
   \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
   \]

4. **Estimate \( n \)**: We need to solve:
   \[
   \frac{n(n+1)(2n+1)}{6} \geq 1996
   \]
   This is a cubic inequality. We can approximate \( n \) by trial and error or using a calculator.

5. **Check values**:
   - For \( n = 19 \):
     \[
     \frac{19 \cdot 20 \cdot 39}{6} = 2470
     \]
     which is greater than 1996.
   - For \( n = 18 \):
     \[
     \frac{18 \cdot 19 \cdot 37}{6} = 2109
     \]
     which is also greater than 1996.
   - For \( n = 17 \):
     \[
     \frac{17 \cdot 18 \cdot 35}{6} = 1785
     \]
     which is less than 1996.

6. **Conclusion**: The smallest \( n \) for which the sum of squares is at least 1996 is \( n = 19 \).

\(\boxed{19}\)