Yes, there is always an equilibrium.

This game is a stochastic game: the number of chips in the pot and the identity of the player whose turn it is to move serve as a state variable. Since the number of chips in the pot is bounded (between 0 and M), there are finitely many states and actions to each player. In fact, the game is a stochastic game with perfect information: the players move alternately, so there are no simultaneous moves. Such games have an equilibrium that do not involve randomization, that is, the choice of the number of chips is deterministic, see Thuijsman and Raghavan, Perfect Information Stochastic Games and Related Classes, International Journal of Game Theory, 1997, 26, 403-408.

Thuijsman and Raghavan (1997) also identify one such equilibrium. Denote by $\sigma_i$ a max-min strategy of player $i$: a deterministic strategy that yields the maximal payoff to player $i$ if all other players play against her (against player $i$) and try to lower her payoff. Denote by $\tau_{-i}$ a deterministic punishment strategy vector of all players except player $i$ who try to lower player $i$'s payoff. Then an equilibrium looks as follows: each player $i$ follows $\sigma_i$. They do so until the first time in which some player, denoted $i_*$, deviates from her $\sigma_{i_*}$. Afterwards, all players punish player $i_*$ -- they switch to $\tau_{-i_*}$.

Edited on 24-July-2021 to reflect the requirement that the equilibrium is in pure stationary strategies.

The game you present is a stochastic game: the number of chips in the pot and the identity of the player whose turn it is to move serve as a state variable. Since the number of chips in the pot is bounded (between 0 and M), there are finitely many states and actions to each player. In fact, the game is a stochastic game with perfect information: the players move alternately, so there are no simultaneous moves.

Such games have (a) an equilibrium that do not involve randomization, that is, the choice of the number of chips is deterministic yet it depends on past play, see Thuijsman and Raghavan, Perfect Information Stochastic Games and Related Classes, International Journal of Game Theory, 1997, 26, 403-408. They also have (b) a symmetric stationary equilibrium that involves randomization, that is, the choice of the number of chips is random and depends only on the current state, see Fink, Equilibrium in a Stochastic n-Person Game, Journal of Science of the Hiroshima University, Series A (mathematics), 1964, 28(1), 89-93.

You, however, are interested in stationary equilibria that involve no randomization. The theory does not guarantee that such equilibria exist.

Yes, there is always an equilibrium.

This game is a stochastic game: the number of chips in the pot and the identity of the player whose turn it is to move serve as a state variable. Since the number of chips in the pot is bounded (between 0 and M), there are finitely many states and actions to each player. In fact, the game is a stochastic game with perfect information: the players move alternately, so there are no simultaneous moves. Such games have an equilibrium that do not involve randomization, that is, the choice of the number of chips is deterministic and depends only on the current state, see Thuijsman and Raghavan, Perfect Information Stochastic Games and Related Classes, International Journal of Game Theory, 1997, 26, 403-408.

Thuijsman and Raghavan (1997) also identify one such equilibrium. Denote by $\sigma_i$ a max-min strategy of player $i$: a deterministic strategy that yields the maximal payoff to player $i$ if all other players play against her (against player $i$) and try to lower her payoff. Denote by $\tau_{-i}$ a deterministic punishment strategy vector of all players except player $i$ who try to lower player $i$'s payoff. Then an equilibrium looks as follows: each player $i$ follows $\sigma_i$. They do so until the first time in which some player, denoted $i_*$, deviates from her $\sigma_{i_*}$. Afterwards, all players punish player $i_*$ -- they switch to $\tau_{-i_*}$.

Yes, there is always an equilibrium.

This game is a stochastic game: the number of chips in the pot and the identity of the player whose turn it is to move serve as a state variable. Since the number of chips in the pot is bounded (between 0 and M), there are finitely many states and actions to each player. In fact, the game is a stochastic game with perfect information: the players move alternately, so there are no simultaneous moves. Such games have an equilibrium that do not involve randomization, that is, the choice of the number of chips is deterministic, see Thuijsman and Raghavan, Perfect Information Stochastic Games and Related Classes, International Journal of Game Theory, 1997, 26, 403-408.

Thuijsman and Raghavan (1997) also identify one such equilibrium. Denote by $\sigma_i$ a max-min strategy of player $i$: a deterministic strategy that yields the maximal payoff to player $i$ if all other players play against her (against player $i$) and try to lower her payoff. Denote by $\tau_{-i}$ a deterministic punishment strategy vector of all players except player $i$ who try to lower player $i$'s payoff. Then an equilibrium looks as follows: each player $i$ follows $\sigma_i$. They do so until the first time in which some player, denoted $i_*$, deviates from her $\sigma_{i_*}$. Afterwards, all players punish player $i_*$ -- they switch to $\tau_{-i_*}$.