Generalized tic-tac-toe We begin with $2n+1$ cards, each with a distinct number from $-n$ to $+n$ on it, face up in between the two players of the game.  The players take turns selecting a card and keeping it.  The first player to collect three cards that sum to zero wins the game.  If the cards are exhausted and neither player has won, a draw is declared.
Tic-tac-toe, or noughts and crosses, is of course the special case $n=4$, by using the essentially unique $3\times3$ magic square:
$$\begin{matrix} 3 & -4 & 1  \\\ -2 & 0 & 2 \\\ -1 & 4& -3\end{matrix}$$
Has the case of general $n$ been studied?
 A: First player wins for $n$ at least five. First turn, name $0$. They name a number, say $-a$. Choose two numbers $b$ and $c$ such that neither $b$, $c$, nor $b+c=a$. Then name $b$, forcing them to name $-b$, then $c$, forcing them to name $-c$, then $-b-c$, winning. You can always choose two such numbers, since each positive number is missed by one of the following triples: $1+2=3, 1+3=4, 1+4=5, 2+3=5$.
As quid points out, this is more complicated than I originally made it seem. If $c\neq a+b$ but $a+b$ is in the interval, then the second player can name $a+b$ in response to $c$ and win.
To avoid this, if $1 <a\leq n-2$, choose $b=1$ and $c=a+1$. Neither $1$, $a+1$, nor $a+2=a$ so this works.
If $a\geq n-1$, choose $b=2$ and $c=1$. Since $n\geq 5$, neither $1$, $2$, nor $3=a$ so this works, and $a+b=a+2>n$.
If $a=1$, choose $b=2$ and $c=3$, so $c=a+b$ and neither $2$, $3$, nor $5=a$.
A: Since I still do not understand the argument for the accepted answer, but agree with its conclusion (win for $n \ge 5$), here is an alternate strategy (albeit not very elegant): 
Let $n \ge 5$.
We start with $0$. And assume without restriction they choose a negative number.
Four cases (but one could somewhat merge 1,3,4): 


*

*They choose $-a$, for $a$ neither $1$, $n-1$, nor $n$. Then, we choose $1$. They have to choose $-1$. We choose $a+1$ defending against their (only) winning move. And then win, since they cannot both 'defend' against $a+1$ and $a+2$ (both being legit due to the condition on $a$).

*They choose $-1$. Then we choose $2$. They need to choose $-2$. We choose $3$ defending against their (only) winning move and creating again two potential wins (at $-3$ and $-4$), and thus winning.

*They choose $-(n-1)$. Then we choose $1$. They have to choose $-1$. We choose $n$ defending their winning move. They need to choose $-n$, which does not create any winning move for them, so we can choose $2$, creating two winning options ($-2$ and $-3$) [note due to $n \ge 5$ there is no interference with the earlier moves], and thus win.

*They chooose $-n$. We choose $1$. They need to choose $-1$. This does not create any threat. So we can choose $2$ and then win with $-2$ or $-3$.    
A: I am not sure about this particular game, but the general and well-studied framework is as follows: given a hypergraph $H$, two players take turns choosing vertices from $H$, the first player collecting a whole edge being the winner. (In your case, the vertex set is $[-n,n]$, and the edges are triples $(a,b,c)\in[-n,n]^3$ which add up to $0$.) Two references you may check: Combinatorial Games: Tic-Tac-Toe Theory and Foundations of Positional Games, both by J. Beck.
