Solution:

The answer is $\mathbf{( C )}$. Observe that for $n \geq 3, a_{n}$ is the number in $\{1, \ldots, 9\}$ given by the remainder of the division by 9 of $a_{n-2}+a_{n-1}$ (where 9 represents remainder 0). In particular, $a_{n-2}$ is the remainder (as above, taken in $\{1, \ldots, 9\}$) in the division by 9 of $a_{n}-a_{n-1}$. Setting $a_{2023}=9$, for each of the 9 possible choices of the value of $a_{2022}$, the value of $a_{2021}$ is determined, and similarly all the other values of the sequence are determined: these 9 choices thus lead us to 9 possible choices of the pair $\left(a_{1}, a_{2}\right)$.