Assume the players have to choose integers, otherwise a best response may not exist. Let the payoff of winning be $\alpha\cdot[\frac23\text{ of the average}]$, $\alpha>0$. Consider the two player (A and B) case, and let's verify whether choosing above $0$ is optimal.
Suppose A chooses $x>0$. Then B can guarantee a win by choosing $x-1$, since $\frac23(x-\frac12)$ is closer to $x-1$ than to $x$. However, if B chooses $0$, A choosing $x=1$ is a best response (A gets zero anyway). So we will have a NE where not both players choose $0$, and the one who chooses $0$ gets positive payoff.
This equilibrium can be generalized to the $n$-player case. Let $n-1$ players choose $0$ and the remaining one choose $1$. In this equilibrium, those who choose $0$ would share the positive payoff (each gets $\frac{\alpha}{n-1}\cdot\frac23\cdot\frac1{n}$). They are best responding because choosing any $x>0$ would imply zero payoff. The one who chooses $1$ is also best responding since he gets zero anyway. (Of course this is not the only equilibrium. There is also one where everyone chooses $0$, which is obvious.)
Admittedly, the above equilibrium relies crucially on the assumption that players are only allowed to choose integers. But this is due to the fact that payoff is a function of the choices (compared to a fixed amount in the original version of the problem). Suppose players are allowed to choose any real number in $[0,100]$. Then if A chooses $x>0$, B would best respond by choosing $x-\epsilon$, where $\epsilon=\min\{y:y>0\}$ (B would win of course, but she also wants to maximize her earning by making her choice as close to A's as possible).
However, such an $\epsilon$ does not exist. Therefore, the only equilibrium is for everyone to choose $0$.
The intuition is this: however large the winning is, you don't get it if you don't win. But winning requires that you choose small. So the incentive to win overwhelms the incentive to win big.
