= Solution
In this <consecutive-number antisymmetric game>, with the row player's number denoted $i$, the <payoff matrix> is
$$
a_{ij}=\begin{cases}
0,&i=j,\\
1,&i=j+1,\\
-1,&j=i+1,\\
-2,&i\ge j+2,\\
2,&j\ge i+2.
\end{cases}
$$
It is an <antisymmetric matrix>, so the <matrix game> has value zero for every $n\ge3$. For $n=3$,
$$
A=\begin{pmatrix}0&-1&2\\1&0&-1\\-2&1&0\end{pmatrix},\qquad
p=q=\left(\frac14,\frac12,\frac14\right)
$$
satisfies $Ap=0$ and $p^TA=0$, proving optimality for both players.
For any larger $n$, extend the same strategy by zero probabilities:
$$
\boxed{v=0,\qquad p=q=\left(\frac14,\frac12,\frac14,0,\ldots,0\right).}
$$
Indeed, the first three coordinates of $Ap$ are zero. For $i=4$, $(Ap)_4=-2/4-2/2+1/4=-5/4$; for $i\ge5$, all three supported numbers are at least two below $i$, so $(Ap)_i=-2$. Hence $Ap\le0$ and, by antisymmetry, $p^TA\ge0$. The column player using $p$ prevents positive payoff, while the row player using $p$ prevents negative payoff, a <mixed-strategy optimality certificate for a matrix game>. This explicitly checks all the additional strategies rather than assuming the three-number subgame remains optimal.
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