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No, there is not. Consider a game with two players, Ann and Bob. Both choose such vectors with entries $0$ or $1$ of the form $(a_1,a_2,\ldots,a_J)$ or $(b_1,b_2,\ldots,b_J)$, respectively. If $\sum_{i=1}^J a_i+b_i$ is odd, Ann wins and Bob loses. If the number is even, Ann loses and Bob wins. Clearly, one of them loses in every profile of pure strategies, ...