Notice that $1000 = 155 \cdot 6 + 7 \cdot 10$. Now we claim that the second player $B$ has the winning strategy: $1)$ As long as there is a $6$ in the above mentioned partition and $A$ takes one of the numbers $1$ to $5$, say $k$, $B$ takes $6 - k$ and we take one $6$ from the partition. $2)$ Whenever $A$ takes $6$, $B$ takes $1$ removing a $7$ from the partition. We see that the number of possible moves taking $6$ is equal to the number of remaining sevens in the partition after $1)$ or $2)$. If during the game all of $6$s are removed before all of the $7$s then if $A$ takes $k$ from $1$ to $6$ than $B$ takes $7 - k$, still keeping the number of possible moves taking $6$ more than or equal to the number of $7$s remaining in the partition, so such a move is always possible. Taking one seven at a time, after $B$ playing $0$ matches will remain and so $B$ wins because after a turn of $A$, the remaining number is never divisible by $7$. Else we are left with a number divisible by $6$ with $A$ playing first so when he takes $k$ from $1$ to $5$, $B$ takes $6 - k$ hence $B$ wins.