Relaxing the notion of Nash Equilibrium Consider a game with $N$ players, each indexed by $i=1,...,N$. Every player $i$ has to choose a $J\times 1$ vector of actions $a_i\equiv (a_{i,1},...,a_{i,J})$ where each $a_{i,j}$ can be zero or one. The payoff of each player $i$ is $u_i(a_i, a_{-i})$, where $a_{-i}$ denotes the actions of the other players.
As a first best, I would like to show that this game has a pure strategy Nash equilibrium (PSNE). I'm unable to do so and I also don't want to allow for mixed strategies. As a second best, I would be happy to consider a weaker notion of equilibrium (still, in pure strategy) and show its existence.
In particular, by revealed preference, if $(a_1,...,a_N)$ is a PSNE, then the following holds:
$$
\forall i=1,...N \quad \forall j=1,...,J: \quad \text{if } a_{i,j}=1\text{, then } u_i(a_i, a_{-i})\geq  u_i(a_{i, \{-j\}}, a_{-i})\\
\hspace{6.5cm} \text{if } a_{i,j}=0\text{, then } u_i(a_{i}, a_{-i})\geq  u_i(a_{i,\{+j\}}, a_{-i})$$
where $a_{i, \{-j\}}$ denotes $a_i$ where $a_{i,j}=1$ is replaced by $a_{i,j}=0$; $a_{i, \{+j\}}$ denotes $a_i$ where $a_{i,j}=0$ is replaced by $a_{i,j}=1$.
Question: Is there any equilibrium notion in pure strategy such that:

it is weaker than PSNE

it implies the revealed preference inequalities reported above

its existence is typically easier to be shown (and, if you could give references on this)
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, and the loser can win by changing only one entry of their chosen vector. So there is no pure strategy profile at which the revealed preference inequalities hold.