Call the strategies of rock, paper, and scissors A, B, and C: C beats B beats A beats C.
Label the possible positions in this game with $2(n-1)$ dollars in the pot as either: $T_{n-1}$ if the previous result was a tie; $G_{n-1}$ if player I has a winning streak of 1 using strategy A (where A could be any of rock, paper or scissors); or $H_{n-1}$ if player I has a winning streak of 2 using strategy A (and will thus take the pot if he wins the next round using strategy A). We are using $n-1$ for the amount in the pot rather than $n$ for algebraic convenience when presenting the sub-game matrices.
(Clearly, this omits the cases where player II has a winning streak, but the values of those positions $G^{-}_{n-1}$ and $H^{-}_{n-1}$ are just the negatives of the values of the corresponding $G_{n-1}$ and $H_{n-1}$ positions.)
Let the value of any position in this game be the optimum game value minus the result that would happen if the players were to quit and immediately split the pot. Then by symmetry
$V(T_{n-1}) = 0$ for all $n >0$ and the optimal strategy for any $T_{n-1}$ position is the trivial one of choosing A, B, or C each with probability 1/3.
Because the value of $T_{n-1}$ is zero, in the position analyses the values and strategies are unaffected if we were to say we don't care how the $2(n-1$ dollars got into the pot, and that any tie ends the game by splitting the pot (and that the players. So the game overall for some value of $n-1$ is characterized by which of 4 positions you are at.
From $G_{n-1}$ the game may end (by a tie, in which case the players restart the game at $T_n$ but since that has zero value we don't care), transition to $H_{n-1}$ with player I gaining a dollar, or transition to $G^{-}_{n-1}$ with player II gaining a dollar.
From $H_{n-1}$ the game may end by a tie, transition to $G_{n-1}$ with player I gaining a dollar having won using strategy B or C, transition to $G^{-}_{n-1}$ with player II gaining a dollar, or end with player I winning the pot (gaining $n$ dollars altogether) by winning using strategy A.
For $n=1$, with no money in he pot, their is no advantage at all to having a winning streak, the position values are all zero, and the optimal strategies are the trivial equal-probability strategies. All the interesting features are in games where there is someting in the pot.
The two game matrices (for $n>1) are:
$$
G_{n-1} = \left(
\begin{array}{ccc}
0 & -g & h \\
+g & 0 & -g \\
-g & g & 0
\end{array}
\right)
$$
(where for convenience we introduce $g \equiv 1+V(G_{n-1})$ and $h \equiv 1+V(H_{n-1})$) and
$$
H_{n-1} = \left(
\begin{array}{ccc}
0 & -g & n \
+g & 0 & -g \
-g & g & 0
\end{array}
\right)
$$
$$
H_{n-1} = \left(
\begin{array}{cc|c}
0 & -g & n \\
+g & 0 & -g \\
-g & g & 0
\end{array}
\right)
$$
To solve $G_{n-1}$ we do the usual mantra of subtracting columns and taking determinants:
$$
G_{n-1}:
\begin{array}{ccc|cc|c}
0 & -g & h & g & -h & 3g^2 \\
+g & 0 & -g & g & 2g & g^2 + 2hg \\
-g & g & 0 & -26 & -g & 2g^2 + hg \\
\hline
-g & g & h+g & & & \\
g & -2g & h & & & \\
\hline
2g^2 + hg & g^2 + 2hg & 3g^2 & & &
\end{array}
$$
He game value is $\frac{hg-g^2}{3h+g6}$. So
$$
g = 1 + V(G_{n-1}) = 1 + \frac{hg-g^2}{3h+g6}
$$
Similarly, we solve $H_{n-1}$:
$$
H_{n-1}:
\begin{array}{ccc|cc|c}
0 & -g & n & g & -n & 3g^2 \\
+g & 0 & -g & g & 2g & g^2 + 2ng \\
-g & g & 0 & -26 & -g & 2g^2 + ng \\
\hline
-g & g & n+g & & & \\
g & -2g & n & & & \\
\hline
2g^2 + ng & g^2 + 2ng & 3g^2 & & &
\end{array}
$$
The game value is $\frac{hg-g^2}{3h+g6}$. So
$$
h = 1 + \frac{ng-g^2}{3n+g6}
$$
Before working with these two equations, we can look at the situation for $n$ very large. There, $ h = 1 + g/3 + O(1/n)$ and we can substitute to find that
$$
\begin{array}{l}
g = \frac{15+\sqrt{1053}}{46} \approx 1.031521 \\
V(G_{\infty}) \approx 0.031521 \\
h = 1 + g/3 \approx 1.34384 \\
V(H_{\infty}) \approx 0.34384
\end{array}
$$
That is, with a very large pot, if player I has a winning streak of 2 wins playing strategy A, player II will very rarely (probability $\frac{3g}{3N+6g}$) risk using strategy C, which could result in losing the whole pot. Player I will almost always choose strategies B (about 2/3 of the time) or C (about 1/3), and the game is favorable to player I with a value of $+\frac{1}{3}$. And if player 1 has just a 1 game winning streak, then the roughly 1.33 reward in the upper right hand corner (going over to $H_{n-1}$ with another A win) biases the game in player I's favor, but only by about 0.03.
Okay, now look at the case for $n$ not large enought to ignore $1/n$ effects: Substituting the expression for $h$ into the expression for $g$ we find
$$
40g^3 + (23n-21)g^2 - (15n+18)g - 9n = 0
$$
So for example, if there is $2 \times 1$ dollars in the pot ($n = 2$) then the value of the game to a player with a winning streak is $G_1 \approx 0.008118$ and the value of a two game winning streak is $G_1 \approx 0.082991$.
In game $G_1$ the strategy for player I is to choose A (the winning streak choice) to (B which beats A) to C in ratio
$$
3g : g + 2h : 2g + h = 3.024 : 3.172 : 3.099
$$
and in game $H_1$ the ratios are
$$
3g : g + 2n : 2g + n = 3.024 : 5.008 : 4.016
$$
The game values for $n = 11$ (ten dollars in the pot) are $G_{10} = 0.024413$ and
$H_{10} = 0.261048$; the strategies for player 1 in game $H_{10}$ are in ratio
$$
3.073 : 21.049: 1.024 \approx 1 : 7 : 4
$$
and for player II they are in ratio
$$
2n+g : n+2g : 3g = 21:024 : 12.049 : 3.073 \approx 7 : 4 : 1
$$
for A, B and C in that order.
Best Answer
Stating the problem
Let $G(n)$ be a rock-paper-scissors game with $n\in\mathbb N$ elements.
Let those $n$ elements be the fist $n$ nonnegative integers $\{0,1,2,\dots,n-1\}=N_n$.
For every $n_0,n_1\in N$, if $n_0$ beats $n_1$, then $n_0$ is a "counter" to $n_1$, and $n_1$ is a "anticounter" to $n_0$. Counters and anticounters of $n_0$ are together called "interactions" of $n_0$.
We want to establish a rule for counters and anticounters of every $n_0\in N_n$, such that:
$n_0$ has an equal number of counters and anticounters - interactions are balanced.
$n_0$'s interactions in $G(n)$ should be kept in $G(n+1)$.
Solution to the problem
The first property requires $n=2k+1,k\in\mathbb N$ to be odd.
If we start observing first few cases of $G(n)$, you'll notice that to satisfy the second property, $n_0=m$ needs to be a counter to $m+1,m-2,m+3,m-4,\dots$
This is finally equivalent to the following rule: The winner between $n_0,n_1\in N_n$ is:
if $n_0,n_1$ have the same parity (both odd or both even), then winner is: $\max\{n_0,n_1\}$
else, if $n_0,n_1$ have $\ne$ parity, then the winner is: $\min\{n_0,n_1\}$
This rule works for defining counters and anticounters for every element.
This rule allows a balanced rock-paper-scissors to be played with any number of odd elements, and to easily add/remove interactions to extend or reduce the game by $2$ elements.
For example, $n=3,5$ cases and with the classic labels to elements $0,1,2,3,4$ :
Generalization to countable sets
Notice that the general observed rule does not depend on $n$.
Thus if we let $n\to\infty$, we can play rock-paper-scissors on $\mathbb N_0 = \mathbb N \cup \{0\}$.
We can also include negative numbers and extend it to the whole integers $\mathbb Z$, as the "parity" and "$\lt$" are still defined on integers.
We can also play it on any countable set. We establish a bijection from some countable set $X$, such as rationals $X=\mathbb Q$ for example, to either $\mathbb N_0$ or $\mathbb Z$, and call it $B(x)$.
Then the winner between $x_1,x_2\in X$ is:
$$=\begin{cases} \displaystyle \max\{B(x_1), B(x_2)\}, & B(x_1)\bmod2=B(x_2)\bmod2\\ \displaystyle \min\{B(x_1), B(x_2)\}, & B(x_1)\bmod2\ne B(x_2)\bmod2 \end{cases}$$
I believe this rule is the most "balanced" way to extend rock-paper-scissors to any finite and countable sets.
Additionally, by symmetry, we can invert the rule and keep all the properties. That is, invert the conditions for min and max calculations.