A bad deal in French Tarot

Let $n$ be the initial number of players, and $k$ the number of cards each player has, except for the last one, who has only $k-2$.
We therefore have:
$$78=(n-1)k+k-2=nk-2$$And hence $nk=80$.
Note that the cards were dealt four at a time, so the number of cards each player has (except for the last one...) is a multiple of 4:
$$k=4t$$(where, incidentally, $t$ is the number of rounds of dealing).
We therefore obtain
$$4nt=80,\quad\text{hence}\quad nt=20.$$

This leaves only the following possibilities:

• $n=20$, $t=1$
• $n=1$, $t=20$ (excluded, since there are several players)
• $n=10$, $t=2$
• $n=2$, $t=10$ (excluded, since the statement specifies that after the player leaves, several players remain).
• $n=5$, $t=4$
• $n=4$, $t=5$.

There is one condition left to consider: since the remaining $n-1$ players divide all the cards of the player who left equally among themselves, this implies that
$$n-1\;\text{divides}\;k-2=4t-2$$

But checking the 4 remaining possibilities, only one of them is compatible with this condition.
Indeed:

• $n=20$, $t=1$ is impossible, since $20-1=19$ does not divide $4\times 1-2=2$.
• $n=10$, $t=2$ is impossible, since $10-1=9$ does not divide $4\times 2-2=6$.
• $n=5$, $t=4$ is impossible, since $5-1=4$ does not divide $4\times 4-2=14$.
• $n=4$, $t=5$ is possible, since $4-1=3$ does indeed divide $4\times 5-2=18$.

Therefore, there could only have been $n=4$ players initially; and hence only 3 players remain.