I think that The simple lottery is an answer to your question. Since I have no background in the game theory, I certainly omit some details and the COMMENTs section below might be far from correct.

On second thought, also The unique-is-the-winner lottery below would be a solution, too.

=THE SIMPLE LOTTERY=

For the purpose of discussion I present more variants of the lottery.

THE GAME. Suppose we have $n=100$ players each of which chooses to buy $x=x_i$ tickets, each of which costs 1 dollar, where

- (v3) $x\in \{0,1,2,3\}$
- (vI) $x\in \{0,1,2,\dots\}$
- (vR) $x$ is a non-negative real number.

The lottery is made with each ticked having equal probability to win the price of $P=202.0202 \dots > 1$ dolars so that $i$-th player wins with probability $x_i/\sum_j x_j$.

ASSUMPTIONS. The number of players is well known. Some game-theory related assumptions might be needed like that the players are inteligent, their goal is to maximize gain expectation (average over probability, in the linear fashion),
the actual move of each player is secret while other information is well known etc. In particular I consider the player's goals uniformity and their public knowledge to be sensitive. (If there are players with different goals like keep-what-I-have-with-certainty or get-maximum-or-noghing, the game might possibly get disbalanced).

THE STRATEGY.
A game-theoretical result like a Nash equilibrium theorem provides us
(I hope; at least in the variant (v3)) with an optimal strategy for each player.
Case (vR) lacks compactness and might be a problem. (vI) is somewhat more compact.

The optimal strategy will be generally a mixed (i.e., probabilistic) strategy.
Because of the symmetry, the mixed strategies of individual players will be the same.

COMMENT.
Though this is not important, I choosed the particular values so that, I guess, (some of) the optimal mixed stragies of the players are actually pure strategies of bying $x=2$ tickets. I think so because assuming(!) the other players play pure strategy of buing $a$ tickets and if I buy $x$ tickets, the outcome expectations is $E(x)=e(x)-x$ where $e(x) = \frac{P x}{(n-1)a + x}$ is win expectation and $x$ is the investment in the tickets.
I consider the variant (vR). The graph of $e(x)$ is an increasing hyperbola with $e(0)=0$ and $\lim_{x\to \infty} e(x) = P = 202.0202\dots$.
If $e'(0)\le 1$ (that is, $a \ge P/(n-1)$) I will buy $x=0$ tickets.
But this is not the equilibrium since the strategy is not optimal to other players
(their expactations are $E_i=P a/((n-1)a) - a = P/(n-1) -a \le 0$ and each of them should consider to buy e.g. $0$ tickets instead).
Otherwise the best pure strategy is to buy $x$ tickets where $e'(x)=1$ ($E'(x)=0$),
which means $x=x_0 = \sqrt{(n-1)a}(\sqrt{P}-\sqrt{(n-1)a})$. For the equilibrium strategy
we have $x_0=a$, which leads to $x_0=a=P(n-1)/n^2$.
If players play like that, the lottery maker gets $P(n-1)/n$ dolars and gives away $P$ dolars, thus paying price $P/n$ for attracting playres to participate in the procedure.
BTW, if $n=1$ and I am the only player, I invest one cent in the tickets and get full $P$.
Now back the mixed strategies: If think of changing my pure $x_0$ ticket strategy to some mixed strategy, I see the utility function is concave which makes the pure $x_0$ strategy
best among all mixed strategies. And more generally, if the strategy of other players is fixed (arbitrarily, mixed), then the utility function is strictly concave (because an average of strictly concave functions), making some pure strategy the optimal strategy. (This even sounds as a proof that the equilibrium strategies for (vR) are pure.) For variants (v3) and (vI), this suggest the optimal would be to mix the two integer values that are next the calculated $x_0$.

=THE ZERO LOTTERY.=

(For the sake of completness I have to note that also the zero lottery seems to be a solution to your problem. The rules are "No one wins." Then, obviously, the chances are equal. That is to clear that you wish to get positive probability that someone wins. May be you wich that probability to be $1$.)

=THE UNIQUE-IS-THE-WINNER LOTTERY=

This lottery has nonzero probabity of no-winner, but it is simpler and more accessible.
Moreover, it can be repeated until a winner is found.

We have $n$ players which have option to by $0$ or $1$ tickets, which costs one dolar. A player wins $P=100$ dolars if he is the only one who bought the ticket. Of no one boght the
ticket, no one wins. If at least two tickets were sould, no one wins.

COMMENT.
Silly player like me chooses to buy a ticket with probability $1/n$. More clever one
perhaps buys with probability proportional to $P$. Even clever player will estimate the price of the game to be about $P/n$, hence, for large $n$, he buys with probability smaller than $P/n$. For smaller $n$ he only observers the lottery is likely to repeat several times and he has no idea what the optimal strategy is.

anytickets. $\endgroup$ – Emil Jeřábek Jan 24 '14 at 16:15