If three numbers are written on the blackboard and one of them is replaced by the (positive) difference of the other two, then after this move one number on the blackboard will be the sum of the other two. Let $a$, $b$ and $a+b$ be the numbers on the blackboard; w.l.o.g. we assume that $b > a$. Because of $a+b-b=a$ and $a+b-a=b$ there is only one possible move. After it the numbers $a$, $b$ and $b-a$ are written on the blackboard. Again, one number (namely $b$) is the sum of the other two and there exists only one possible move.
This means that at the latest from the second turn on there is no choice of moves and all moves are inevitable. Furthermore, from the second move on, the largest of the three numbers is decreased, and since no number can become negative, after a finite number of moves one of the numbers will be $0$. Since $0$ is the difference of the other two numbers, we must have $0$, $a$, $a$ on the blackboard. Now $a-0=a$ and $a-a=0$, therefore no further move is possible. Thus the player writing $0$, $a$, $a$ onto the blackboard is the winner.
If the game starts with the numbers $2000$, $17$ and $2017$ on the blackboard, the course of the game is as follows:
1st move (A): $2000$, $17$, $1983$
2nd move (B): $1966$, $17$, $1983$
3rd move (A): $1966$, $17$, $1949$
etc. (since $2000 : 17 = 117.6$ ...)
117th move (A): $2000 - 116 \cdot 17 = 28$, $17$, $2000 - 117 \cdot 17 = 11$
118th move (B): $6$, $17$, $11$
119th move (A): $6$, $5$, $11$
120th move (B): $6$, $5$, $1$
121st move (A): $4$, $5$, $1$
122nd move (B): $4$, $3$, $1$
123rd move (A): $2$, $3$, $1$
124th move (B): $2$, $1$, $1$
125th move (A): $0$, $1$, $1$
and A wins the game.