Answer. $100$.

Solution 1. We prove a more general statement for sets of cardinality $n$ (the problem being the special case $n=100$, then the answer is $n$). In the following, we write $A>B$ or $B<A$ for "$A$ beats $B$".

Part I. Let us first define $n$ different rules that satisfy the conditions. To this end, fix an index $k \in \{1,2, \ldots, n\}$. We write both $A$ and $B$ in increasing order as $A=\{a_{1}, a_{2}, \ldots, a_{n}\}$ and $B=\{b_{1}, b_{2}, \ldots, b_{n}\}$ and say that $A$ beats $B$ if and only if $a_{k}>b_{k}$. This rule clearly satisfies all three conditions, and the rules corresponding to different $k$ are all different. Thus there are at least $n$ different rules.

Part II. Now we have to prove that there is no other way to define such a rule. Suppose that our rule satisfies the conditions, and let $k \in \{1,2, \ldots, n\}$ be minimal with the property that
$$A_{k}=\{1,2, \ldots, k, n+k+1, n+k+2, \ldots, 2n\} \prec B_{k}=\{k+1, k+2, \ldots, n+k\}.$$
Clearly, such a $k$ exists, since this holds for $k=n$ by assumption. Now consider two disjoint sets $X=\{x_{1}, x_{2}, \ldots, x_{n}\}$ and $Y=\{y_{1}, y_{2}, \ldots, y_{n}\}$, both in increasing order (i.e., $x_{1}<x_{2}<\cdots<x_{n}$ and $y_{1}<y_{2}<\cdots<y_{n}$). We claim that $X<Y$ if (and only if - this follows automatically) $x_{k}<y_{k}$.

To prove this statement, pick arbitrary real numbers $u_{i}, v_{i}, w_{i} \notin X \cup Y$ such that
$$u_{1}<u_{2}<\cdots<u_{k-1}<\min(x_{1}, y_{1}), \quad \max(x_{n}, y_{n})<v_{k+1}<v_{k+2}<\cdots<v_{n},$$
and
$$x_{k}<v_{1}<v_{2}<\cdots<v_{k}<w_{1}<w_{2}<\cdots<w_{n}<u_{k}<u_{k+1}<\cdots<u_{n}<y_{k},$$
and set
$$U=\{u_{1}, u_{2}, \ldots, u_{n}\}, V=\{v_{1}, v_{2}, \ldots, v_{n}\}, W=\{w_{1}, w_{2}, \ldots, w_{n}\}.$$
Then
- $u_{i}<y_{i}$ and $x_{i}<v_{i}$ for all $i$, so $U<Y$ and $X<V$ by the second condition.
- The elements of $U \cup W$ are ordered in the same way as those of $A_{k-1} \cup B_{k-1}$, and since $A_{k-1}>B_{k-1}$ by our choice of $k$, we also have $U>W$ (if $k=1$, this is trivial).
- The elements of $V \cup W$ are ordered in the same way as those of $A_{k} \cup B_{k}$, and since $A_{k}<B_{k}$ by our choice of $k$, we also have $V<W$.
It follows that
$$X<V<W<U<Y$$
so $X<Y$ by the third condition, which is what we wanted to prove.

Solution 2. Another possible approach to Part II of this problem is induction on $n$. For $n=1$, there is trivially only one rule in view of the second condition.

In the following, we assume that our claim (namely, that there are no possible rules other than those given in Part I) holds for $n-1$ in place of $n$. We start with the following observation:

Claim. At least one of the two relations
$$(\{2\} \cup \{2i-1 \mid 2 \leqslant i \leqslant n\}) \prec (\{1\} \cup \{2i \mid 2 \leqslant i \leqslant n\})$$
and
$$(\{2i-1 \mid 1 \leqslant i \leqslant n-1\} \cup \{2n\}) \prec (\{2i \mid 1 \leqslant i \leqslant n-1\} \cup \{2n-1\})$$
holds.

Proof. Suppose that the first relation does not hold. Since our rule may only depend on the relative order, we must also have
$$(\{2\} \cup \{3i-2 \mid 2 \leqslant i \leqslant n-1\} \cup \{3n-2\}) > (\{1\} \cup \{3i-1 \mid 2 \leqslant i \leqslant n-1\} \cup \{3n\}).$$
Likewise, if the second relation does not hold, then we must also have
$$(\{1\} \cup \{3i-1 \mid 2 \leqslant i \leqslant n-1\} \cup \{3n\}) > (\{3\} \cup \{3i \mid 2 \leqslant i \leqslant n-1\} \cup \{3n-1\}).$$
Now condition 3 implies that
$$(\{2\} \cup \{3i-2 \mid 2 \leqslant i \leqslant n-1\} \cup \{3n-2\}) > (\{3\} \cup \{3i \mid 2 \leqslant i \leqslant n-1\} \cup \{3n-1\}),$$
which contradicts the second condition.

Now we distinguish two cases, depending on which of the two relations actually holds:

First case: $(\{2\} \cup \{2i-1 \mid 2 \leqslant i \leqslant n\}) \prec (\{1\} \cup \{2i \mid 2 \leqslant i \leqslant n\})$.
Let $A=\{a_{1}, a_{2}, \ldots, a_{n}\}$ and $B=\{b_{1}, b_{2}, \ldots, b_{n}\}$ be two disjoint sets, both in increasing order. We claim that the winner can be decided only from the values of $a_{2}, \ldots, a_{n}$ and $b_{2}, \ldots, b_{n}$, while $a_{1}$ and $b_{1}$ are actually irrelevant. Suppose that this was not the case, and assume without loss of generality that $a_{2}<b_{2}$. Then the relative order of $a_{1}, a_{2}, \ldots, a_{n}, b_{2}, \ldots, b_{n}$ is fixed, and the position of $b_{1}$ has to decide the winner. Suppose that for some value $b_{1}=x$, $B$ wins, while for some other value $b_{1}=y$, $A$ wins.

Write $B_{x}=\{x, b_{2}, \ldots, b_{n}\}$ and $B_{y}=\{y, b_{2}, \ldots, b_{n}\}$, and let $\varepsilon>0$ be smaller than half the distance between any two of the numbers in $B_{x} \cup B_{y} \cup A$. For any set $M$, let $M \pm \varepsilon$ be the set obtained by adding/subtracting $\varepsilon$ to all elements of $M$. By our choice of $\varepsilon$, the relative order of the elements of $(B_{y}+\varepsilon) \cup A$ is still the same as for $B_{y} \cup A$, while the relative order of the elements of $(B_{x}-\varepsilon) \cup A$ is still the same as for $B_{x} \cup A$. Thus $A \prec B_{x}-\varepsilon$, but $A \succ B_{y}+\varepsilon$. Moreover, if $y>x$, then $B_{x}-\varepsilon < B_{y}+\varepsilon$ by condition 2, while otherwise the relative order of the elements in $(B_{x}-\varepsilon) \cup (B_{y}+\varepsilon)$ is the same as for the two sets $\{2\} \cup \{2i-1 \mid 2 \leqslant i \leqslant n\}$ and $\{1\} \cup \{2i \mid 2 \leqslant i \leqslant n\}$, so that $B_{x}-\varepsilon < B_{y}+\varepsilon$. In either case, we obtain
$$A \prec B_{x}-\varepsilon \prec B_{y}+\varepsilon \prec A$$
which contradicts condition 3.

So we know now that the winner does not depend on $a_{1}, b_{1}$. Therefore, we can define a new rule $<^{*}$ on sets of cardinality $n-1$ by saying that $A<^{*} B$ if and only if $A \cup \{a\} < B \cup \{b\}$ for some $a, b$ (or equivalently, all $a, b$) such that $a<\min A$, $b<\min B$ and $A \cup \{a\}$ and $B \cup \{b\}$ are disjoint. The rule $<^{*}$ satisfies all conditions again, so by the induction hypothesis, there exists an index $i$ such that $A<^{*} B$ if and only if the $i^{\text{th}}$ smallest element of $A$ is less than the $i^{\text{th}}$ smallest element of $B$. This implies that $C<D$ if and only if the $(i+1)^{\text{th}}$ smallest element of $C$ is less than the $(i+1)^{\text{th}}$ smallest element of $D$, which completes our induction.

Second case: $(\{2i-1 \mid 1 \leqslant i \leqslant n-1\} \cup \{2n\}) \prec (\{2i \mid 1 \leqslant i \leqslant n-1\} \cup \{2n-1\})$.
Set $-A=\{-a \mid a \in A\}$ for any $A \subseteq \mathbb{R}$. For any two disjoint sets $A, B \subseteq \mathbb{R}$ of cardinality $n$, we write $A \prec^{\circ} B$ to mean $(-B)<(-A)$. It is easy to see that $\prec^{\circ}$ defines a rule to determine a winner that satisfies the three conditions of our problem as well as the relation of the first case. So it follows in the same way as in the first case that for some $i$, $A \prec^{\circ} B$ if and only if the $i^{\text{th}}$ smallest element of $A$ is less than the $i^{\text{th}}$ smallest element of $B$, which is equivalent to the condition that the $i^{\text{th}}$ largest element of $-A$ is greater than the $i^{\text{th}}$ largest element of $-B$. This proves that the original rule $\prec$ also has the desired form.