, 2022

Since $1 \le a_1 < a_2 < \dots < a_{2022}$, we have $a_i \ge i$ for all $1 \le i \le 2022$. Also, the condition $a_1^2 - 6^2 \ge a_2^2 - 7^2 \ge \dots \ge a_{2022}^2 - 2027^2$ is equivalent to $a_{i+1}^2 - a_i^2 \le (i+6)^2 - (i+5)^2 = 2i + 11$ ($1 \le i \le 2021$).

When $a_{i+1} \ge a_i + 3$ holds for some $1 \le i \le 2021$, we have $a_{i+1}^2 - a_i^2 \ge (a_i + 3)^2 - a_i^2 = 6a_i + 9 \ge 6i + 9 > 2i + 11$ for such $i$, which contradicts to $a_{i+1}^2 - a_i^2 \le 2i + 11$. Therefore, since $a_i < a_{i+1}$, we have $a_{i+1} = a_i + 1$ or $a_{i+1} = a_i + 2$ ($1 \le \forall i \le 2021$).
For each $i$, if $a_{i+1} = a_i + 1$, then since $a_{i+1}^2 - a_i^2 = 2a_i + 1$, $a_{i+1}^2 - a_i^2 \le 2i + 11$ is equivalent to $a_i \le i + 5$. Therefore, the tuples satisfying $a_{i+1} = a_i + 1$ for all $1 \le i \le 2021$ are of the form $a_i = i + c$ for a constant integer $0 \le c \le 5$. There are exactly six such tuples.
Hereafter, we assume that $a_{i+1} = a_i + 2$ for some $1 \le i \le 2021$ and denote the maximum of such $i$ by $j$. Since $4j + 4 \le 4a_j + 4 = a_{j+1}^2 - a_j^2 \le 2j + 11$, we have $j = 1, 2, 3$.
* When $j = 1$, since $4a_1 + 4 = a_2^2 - a_1^2 \le 2 \cdot 1 + 11$, we have $a_1 = 1, 2$. In this case, $(a_1, a_2, \dots, a_{2022}) = (1, 3, 4, 5, \dots, 2022, 2023)$, $(2, 4, 5, 6, \dots, 2023, 2024)$.
* When $j = 2$, since $4a_2 + 4 = a_3^2 - a_2^2 \le 2 \cdot 2 + 11$ and $a_2 \ge 2$, we have $a_2 = 2$. In this case, $(a_1, a_2, \dots, a_{2022}) = (1, 2, 4, 5, 6, \dots, 2022, 2023)$.
* When $j = 3$, since $4a_3 + 4 = a_4^2 - a_3^2 \le 2 \cdot 3 + 11$ and $a_3 \ge 3$, we have $a_3 = 3$. In this case, $(a_1, a_2, \dots, a_{2022}) = (1, 2, 3, 5, 6, 7, \dots, 2022, 2023)$.
Hence the total number of the tuples is $6 + 4 = 10$.