Two sequences $\{a_i\}$ and $\{b_i\}$ are defined as follows: $\{ a_i \} = 0, 3, 8, \dots, n^2 - 1, \dots$ and $\{ b_i \} = 2, 5, 10, \dots, n^2 + 1, \dots $. If both sequences are defined with $i$ ranging across the natural numbers, how many numbers belong to both sequences?

[i]Proposed by Isabella Grabski[/i]
1. We start by analyzing the sequences $\{a_i\}$ and $\{b_i\}$:
   - The sequence $\{a_i\}$ is given by $a_i = n^2 - 1$ for $n \in \mathbb{N}$.
   - The sequence $\{b_i\}$ is given by $b_i = n^2 + 1$ for $n \in \mathbb{N}$.

2. Suppose there is a number $k$ that belongs to both sequences. Then there exist integers $a$ and $b$ such that:
   \[
   k = a^2 - 1 = b^2 + 1
   \]
   This can be rewritten as:
   \[
   a^2 - 1 = b^2 + 1
   \]
   Simplifying, we get:
   \[
   a^2 - b^2 = 2
   \]

3. We can factorize the left-hand side using the difference of squares:
   \[
   (a + b)(a - b) = 2
   \]

4. Now, we need to find integer solutions for $(a + b)$ and $(a - b)$ such that their product is 2. The possible pairs $(a + b, a - b)$ are:
   \[
   (1, 2) \quad \text{and} \quad (2, 1)
   \]
   However, both pairs must have the same parity (both even or both odd) because $a$ and $b$ are integers. 

5. Let's analyze the parity:
   - If $a + b$ and $a - b$ are both odd, their product would be odd, which contradicts the fact that their product is 2 (an even number).
   - If $a + b$ and $a - b$ are both even, their product would be divisible by 4, which again contradicts the fact that their product is 2.

6. Since neither of the conditions can be satisfied, there are no integer solutions for $a$ and $b$ such that $a^2 - b^2 = 2$.

7. Therefore, there are no numbers that belong to both sequences $\{a_i\}$ and $\{b_i\}$.

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\boxed{0}
\]