Find the maximum number of pairwise disjoint sets of the form
$S_{a,b} = \{n^{2}+an+b | n \in \mathbb{Z}\}$, $a, b \in \mathbb{Z}$.
1. **Observation on the form of sets $S_{a,b}$:**
   - We start by noting that the sets $S_{a,b} = \{n^2 + an + b \mid n \in \mathbb{Z}\}$ can be simplified based on the parity of $a$.
   - If $a$ is even, let $a = 2k$. Then:
     \[
     S_{a,b} = \{n^2 + 2kn + b \mid n \in \mathbb{Z}\}
     \]
     We can rewrite this as:
     \[
     n^2 + 2kn + b = (n+k)^2 + b - k^2
     \]
     Hence, $S_{a,b} = S_{0, b - k^2}$.
   - If $a$ is odd, let $a = 2k + 1$. Then:
     \[
     S_{a,b} = \{n^2 + (2k+1)n + b \mid n \in \mathbb{Z}\}
     \]
     We can rewrite this as:
     \[
     n^2 + (2k+1)n + b = (n+k)^2 + (n+k) + b - k^2 - k
     \]
     Hence, $S_{a,b} = S_{1, b - k^2 - k}$.

2. **Reduction to simpler families:**
   - Based on the above observations, we can reduce the problem to considering sets of the form $S_{0,m}$ and $S_{1,m}$ for some integer $m$.

3. **Claim 1: For any $m$ and $m'$, $S_{1,m} \cap S_{0,m'} \neq \emptyset$:**
   - We need to find integers $n_0$ and $n_1$ such that:
     \[
     n_0^2 + n_0 + m = n_1^2 + m'
     \]
   - Multiplying both sides by 4 and completing the square:
     \[
     4n_0^2 + 4n_0 + 4m + 1 = 4n_1^2 + 4m' + 1
     \]
     \[
     (2n_0 + 1)^2 + 4m = (2n_1)^2 + 4m'
     \]
     \[
     (2n_0 + 1 - 2n_1)(2n_0 + 1 + 2n_1) = 4m' - 4m + 1
     \]
   - Let $2n_0 + 1 - 2n_1 = 1$ (which gives $n_0 = n_1$), and $4n_0 + 1 = 4m' - 4m + 1$. This implies:
     \[
     n_0 = m' - m
     \]
   - Hence, for this pair, the two given sets intersect.

4. **Claim 2: Let $I_0$ and $I_1$ be two index sets such that:**
   - For $m, m' \in I_0$, $m \neq m' \implies S_{0,m} \cap S_{0,m'} = \emptyset$.
   - For $n, n' \in I_1$, $n \neq n' \implies S_{1,n} \cap S_{1,n'} = \emptyset$.
   - We need to show that $|I_0|, |I_1| \leq 2$.

5. **Proof for $I_0$:**
   - We want to ensure that no pair $(x, y)$ satisfies:
     \[
     x^2 + m = y^2 + m'
     \]
     \[
     (x - y)(x + y) = m' - m
     \]
   - If $m' - m$ is odd, then the pair:
     \[
     (x, y) = \left(\frac{m' - m + 1}{2}, \frac{m' - m - 1}{2}\right)
     \]
     achieves the equality.
   - If $4 \mid m' - m$, then letting $x - y = 2$ and $x + y = \frac{m' - m}{2}$, we get:
     \[
     (x, y) = \left(\frac{m' - m}{4} + 1, \frac{m' - m}{4} - 1\right)
     \]
     achieves the equality.
   - The only remaining case is $m' - m \equiv 2 \pmod{4}$. In this case, there is no solution since $x - y \equiv x + y \pmod{2}$, making the product divisible by 4, which is a contradiction.
   - If $|I_0| \geq 3$, take three distinct members $m, m'$, and $m''$ of $I_0$. From the given conditions, $m - m' \equiv 2 \pmod{4}$ and $m - m'' \equiv 2 \pmod{4}$. Subtraction gives $m' - m'' \equiv 0 \pmod{4}$, hence $S_{0,m'} \cap S_{0,m''} \neq \emptyset$, contradiction.

6. **Proof for $I_1$:**
   - We want to ensure that no pair $(x, y)$ solves:
     \[
     x^2 + x + n = y^2 + y + n'
     \]
     \[
     (x - y)(x + y + 1) = n' - n
     \]
   - Note that $x - y$ and $x + y + 1$ have opposite parity modulo 4. If $n' - n$ is not divisible by 2, this equation has no solutions.
   - If $n' - n \equiv 2 \pmod{4}$, letting $n' - n = 4u + 2$, $x - y = 2$, and $x + y = 2u$, we get:
     \[
     (x, y) = \left(u + 1, u - 1\right)
     \]
     achieves the pair.
   - If $4 \mid n' - n$, letting $n' - n = 4u$, we can let $x - y = 1$ and $x + y + 1 = 4u$, solving as:
     \[
     (x, y) = \left(2u, 2u - 1\right)
     \]
     achieves the pair.
   - If $|I_1| \geq 3$, take three members $n, n'$, and $n''$. We get $n - n'$ is odd, and $n' - n''$ is odd, hence $n - n''$ must be even, making $S_{1,n} \cap S_{1,n''} \neq \emptyset$, contradiction.

7. **Conclusion:**
   - Thus, at most two such sets can be taken.

The final answer is $\boxed{2}$