Answer: $n=14$ works . First I'll show that $n=13$ doesen't work. Let the first player take card with number $4$ first. If the second player takes one of the pair $(8,9) $ or $(9,10)$ then the first player takes $(10,11,12)$ and $(5,6,7) $ respectively ,and it can't take 4 consecutive numbers. If the second player doesn't take $ (8,9)$ or $(9,10)$ then the first takes $(8,9,10)$ and then the second player is left with $(1,2,3) , (5,6,7) $ and $(11,12,13)$ ,so it can't take $4$ consecutive numbers. To show that $n=14$ works : Let the first player take card with number $k$ on his move . We have on one hand $k-1$ consecutive numbers , and the other $ 14-k$ . W.L.O.G let $k-1 \leq 14-k \Longleftrightarrow k \leq 7$ If $k-1 \geq 4$ then the second player takes $(k+1,k+2)$ , and now there are 2 groups $(1,2,3,4,\dots ,k) $ and $(k+3,k+4,\dots,14)$ each with more than $4$ elements , so second player can always take $4$ consecutive numbers. If $k=4$ or $3 $ then second player takes $(1,2)$ and if $k=1,2$ then second player takes $(3,4)$ and in both cases we are left with at least with $10$ consecutive numbers , and no matter of how one takes $3$ consecutive numbers , there are still left 4 consecutive numbers . $\blacksquare$