Let Mary always choose the card with the maximal number among the two leftmost cards. Then on $i$-th move the difference between the sums of numbers in her pockets increases at least by $1$ which implies that this strategy allows to get at least $S = 2 + 4 + \dots + 2022 = 1011 \cdot 1012$. In particular, $S(P) \ge M$ for each $P$ and moreover if at some moment the difference between the cards is more than $1$ then Mary gets the sum more than $M$. Hence if $S(P) = M$ then the row consists of the pairs $\{1, 2\}$, $\{3, 4\}$, $\dots$, $\{2021, 2022\}$ arranged in some order.

Suppose that $S(P) = M$. If there exist two adjacent pairs $\{2i - 1, 2i\}$ and $\{2j - 1, 2j\}$ (in that order from left to right) such that $i < j$ then choosing two numbers from the second pair instead of the first pair will allow Mary to obtain the sum greater than $M$, a contradiction. Hence if $i < j$ any pair $\{2j - 1, 2j\}$ is placed to the left from any pair $\{2i - 1, 2i\}$.

Finally consider any such initial position $P$, it has the form $\{2021, 2022\}$, $\{2019, 2020\}$, $\ldots$, $\{1, 2\}$ where the elements in the pairs are arbitrarily permuted. On $i$-th move Mary will put into her left pocket the card with number which is at least $2023 - 2i$. Hence the sum of the numbers in her left pocket will be not less than $2021 + 2019 + \ldots + 1$, which implies that the sum in her right pocket will not exceed $M$. Therefore $S(P) = M$ for all $2^{1011}$ such positions.