For two sets $A, B$, define the operation $$A \otimes B = \{x \mid x=ab+a+b, a \in A, b \in B\}.$$ Set $A=\{0, 2, 4, \cdots, 18\}$ and $B=\{98, 99, 100\}$. Compute the sum of all the elements in $A \otimes B$.

[i](Source: China National High School Mathematics League 2021, Zhejiang Province, Problem 7)[/i]
1. First, we start by understanding the operation defined for sets \( A \) and \( B \):
   \[
   A \otimes B = \{x \mid x = ab + a + b, \, a \in A, \, b \in B\}
   \]
   We can rewrite \( x = ab + a + b \) as:
   \[
   x = (a+1)(b+1) - 1
   \]

2. Given sets \( A = \{0, 2, 4, \ldots, 18\} \) and \( B = \{98, 99, 100\} \), we need to compute the sum of all elements in \( A \otimes B \).

3. We first calculate the sum of the elements in \( A \):
   \[
   A = \{0, 2, 4, \ldots, 18\}
   \]
   This is an arithmetic sequence with the first term \( a_1 = 0 \), common difference \( d = 2 \), and the last term \( a_n = 18 \). The number of terms \( n \) can be found by:
   \[
   a_n = a_1 + (n-1)d \implies 18 = 0 + (n-1) \cdot 2 \implies n = 10
   \]
   The sum of the first \( n \) terms of an arithmetic sequence is given by:
   \[
   S_n = \frac{n}{2} (a_1 + a_n) = \frac{10}{2} (0 + 18) = 5 \cdot 18 = 90
   \]

4. Next, we calculate the sum of all elements in \( A \otimes B \):
   \[
   A \otimes B = \{(a+1)(b+1) - 1 \mid a \in A, b \in B\}
   \]
   We need to compute the sum for each \( b \in B \):
   \[
   \sum_{a \in A} \sum_{b \in B} ((a+1)(b+1) - 1)
   \]

5. We first compute the inner sum for a fixed \( b \):
   \[
   \sum_{a \in A} ((a+1)(b+1) - 1) = (b+1) \sum_{a \in A} (a+1) - \sum_{a \in A} 1
   \]
   We already know \( \sum_{a \in A} a = 90 \). Now, we compute \( \sum_{a \in A} (a+1) \):
   \[
   \sum_{a \in A} (a+1) = \sum_{a \in A} a + \sum_{a \in A} 1 = 90 + 10 = 100
   \]
   Therefore:
   \[
   \sum_{a \in A} ((a+1)(b+1) - 1) = (b+1) \cdot 100 - 10 = 100b + 100 - 10 = 100b + 90
   \]

6. Now, we sum this result over all \( b \in B \):
   \[
   \sum_{b \in B} (100b + 90) = 100 \sum_{b \in B} b + 90 \sum_{b \in B} 1
   \]
   \[
   \sum_{b \in B} b = 98 + 99 + 100 = 297
   \]
   \[
   \sum_{b \in B} 1 = 3
   \]
   Therefore:
   \[
   \sum_{b \in B} (100b + 90) = 100 \cdot 297 + 90 \cdot 3 = 29700 + 270 = 29970
   \]

7. Finally, the sum of all elements in \( A \otimes B \) is:
   \[
   \boxed{29970}
   \]