28. Two players, A and B, place a chess piece at each of two points on a straight line, with a distance of $L (L$ being a positive integer $)$ between them. A and B take turns to move their chess pieces along the line in any chosen direction, with movement distances of $a, b (a < b, a, b$ being positive integers), respectively. When the two chess pieces meet, A loses. Determine the necessary and sufficient conditions for B to have a winning strategy.
28. The necessary and sufficient condition for player B to win is $b=2a$ and $L$ is an integer multiple of $a$. (1) Necessity. If $b \neq 2a$, then as long as $L \neq 0$, there will be at least one of the two numbers $L+a$ and $|L-a|$ that is neither 0 nor $b$ (if $|L-a|=0$, then $L+a=2a \neq b$ and $L+a \neq 0$; if $|L-a|=b$, then $L+a \neq b$ and $L+a \neq 0$). Therefore, player A can always move the chess piece so that the distance between the two pieces is neither 0 nor $b$, thus preventing player B from winning on the next move. If $b=2a$, but $L$ is not a positive integer multiple of $a$, then since the distances moved by players A and B are $a$ and $b=2a$ respectively, which are both multiples of $a$, it is impossible to make the distance between the two pieces 0, meaning player B cannot win. The necessity is thus proven. (2) Sufficiency. Suppose $b=2a$ and $L$ is a positive integer multiple of $a$. If the distance between the two pieces is greater than $a$, no matter how player A moves the piece, player B can always make the distance between the two pieces decrease by $b-a=a$. After a finite number of moves, player B can always make the distance between the two pieces $d \leqslant a$. Since $L$ is a positive integer multiple of $a$, the distance between the two pieces will always decrease by $a$ after each turn of players A and B, so $d$ must be an integer multiple of $a$, hence $d=0$ or $a$. If $d=0$, then player B wins. If $d=a$, then player A can only move the pieces to a distance of 0 (at which point player B has already won) or to $2a=b$. At this point, player B can move once to make the distance between the two pieces 0 and win, thus proving the sufficiency.