Example 1 Given 20 numbers $1,2, \cdots, 20$, two people, A and B, take turns placing a “+” or “-” sign in front of these numbers (the order of placement is not limited). After 20 steps, the absolute value $S$ of the resulting algebraic sum is calculated. A wants to make $S$ as small as possible, while B wants to make $S$ as large as possible. What is the largest $S$ that B can guarantee?
Divide the numbers into 10 pairs: $(1,2),(3,4), \cdots,(19,20)$. Player B first ensures that 19 and 20 have the same sign. Due to Player A's interference, it is not possible to guarantee that the other pairs have the same sign. When Player A adds a sign in front of a number, Player B adds a sign in front of the other number in the same pair. In the first 9 pairs $(1,2),(3,4), \cdots,(17,18)$, Player B uses the opposite sign of Player A, and in the last pair $(19,20)$, Player B uses the same sign as Player A. This ensures that $S \geqslant 20+19-9 \times 1=30$.

On the other hand, Player A's strategy is to counteract. Let $S_{i}$ be the algebraic sum of the $2i$ numbers that have already been assigned signs after Player B's $i$-th step. Then, Player A places a sign opposite to $S_{i}$ in front of the largest remaining number $a_{i+1}$ (if $S_{i}=0$, the sign can be placed arbitrarily). Suppose Player A last changes the sign of the algebraic sum at his $k$-th step $(1 \leqslant k \leqslant 10)$, i.e., $a_{k} \geqslant\left|S_{k-1}\right|$. At this point,
$\left|S_{k-1} \pm a_{k}\right| \leqslant a_{k} \leqslant 21-k$ (since $20=a_{1}>a_{2}>\cdots>a_{k-1}>a_{k}$, so $a_{k} \leqslant 20-(k-1)=$ $21-k)$, and the number $b_{k}$ in front of which Player B places a sign at his $k$-th step is $\leqslant a_{k}-1 \leqslant 20-k$, hence
$$
\left|S_{k}\right|=\left|S_{k-1} \pm a_{k} \pm b_{k}\right| \leqslant\left|S_{k-1} \pm a_{k}\right|+b_{k} \leqslant 41-2 k \text {. }
$$

Afterwards, in each round, $b_{j}<a_{j}<\left|S_{j-1}\right|$, so $\left|S_{j}\right| \leqslant\left|S_{j-1}\right|-\left|a_{j}-b_{j}\right| \leqslant\left|S_{j-1}\right|-1$. Therefore, the final result is
$$
S=\left|S_{10}\right|=\left|S_{k}\right|+\sum_{j=k+1}^{10}\left(\left|S_{j}\right|-\left|S_{j-1}\right|\right) \leqslant 41-2 k-(10-k)=31-k \leqslant 30 .
$$

If Player A cannot change the sign of the algebraic sum in each round, then in each round, $\left|S_{j}\right| \leqslant\left|S_{j-1}\right|-1$, so
$$
S \leqslant a_{1}+b_{1}-9 \leqslant 20+19-9=30 \text {. }
$$

Therefore, the maximum $S$ that Player B can ensure is 30.