For $i = 1, 2, \ldots, N$, let $X_i$ be the number of X contained in $S_i$, and $Y_i$ be the sum of digits contained in $S_i$, regarding them as one-digit integers.

Choose an arbitrary permutation $P = (P_1, P_2, \ldots, P_N)$ of $(1, 2, \ldots, N)$ and an integer $1 \leq i \leq N-1$. Consider a sequence that is obtained by swapping the $i$-th and $(i+1)$-th element of $P$, $P' = (P_1, P_2, \ldots, P_{i-1}, P_{i+1}, P_i, P_{i+2}, \ldots, P_N)$. Let us compare the score $f(P)$ of $T$ corresponding to $P$, and the score $f(P')$ of $T$ corresponding to $P'$.

As a result of the update from $P$ to $P'$, $T$ loses “the pairs of an X contained in $S_{P_i}$ and a digit contained in $S_{P_{i+1}}$,” and gains “the pairs of an X contained in $S_{P_{i+1}}$ and a digit contained in $S_{P_i}$.” Thus, $f(P') - f(P) = X_{P_{i+1}}Y_{P_i} - X_{P_i}Y_{P_{i+1}}$. Therefore, we have

$$f(P') \gt f(P)$$

$$\Leftrightarrow X_{P_{i+1}}Y_{P_i} \gt X_{P_i}Y_{P_{i+1}} \tag{1}$$

$$\Leftrightarrow \frac{Y_{P_i}}{X_{P_i}} \gt \frac{Y_{P_{i+1}}}{X_{P_{i+1}}}. \tag{2}$$

In other words, swapping the $i$-th and $(i+1)$-th elements of $P$ makes the score of $T$ larger if and only if the inequality (2) holds.

Therefore, in order to maximize the score of $T$, let $P$ be a permutation of $(1, 2, \ldots, N)$ sorted in ascending order of $(1, 2, \ldots, N)$. (Here, if $X_i= 0$, we consider that $Y_i/X_i$ takes a value $\infty$ greater than any real number.)

Hence, in order to solve this problem, it is sufficient to construct $T$ by concatenating $S_1, S_2, \ldots, S_N$ in ascending order of $Y_i/X_i$ and to print the score of that $T$. In the implementation, it would be simple to implement the inequality (1) instead of (2), because we can stick to integers and handle the exception where the denominator is $0$.