Irish Mathematical Olympiad

Suppose that each pair of club members participates together in $r$ walks. The total number $m$ of walks organised by the club is then given by
$$m\binom{2n}{2} = r\binom{4n}{2},$$
since the right-hand side counts the total number of walks by considering all pairs of club members and using rule (d), but counts each walk $\binom{2n}{2}$ times. Simplifying, we obtain $2r(4n-1) = m(2n-1)$. Since $2n-1$ is relatively prime to $2$ and $4n-1$, we must have $2n-1 \mid r$, i.e. $r = (2n-1)p$, and so $m = 2(4n-1)p$ (for some positive integer $p$). From rules (b) and (c), each club member takes part in exactly $m/2 = (4n-1)p$ walks.

Next, let $S_i$ denote the set of walks in which club member $i$ participates, let $S_{i,j}$ denote the set of walks in which club members $i$ and $j$ participate, and let $S_{i,j,k}$ denote the set of walks in which club members $i$, $j$ and $k$ participate. Then we have $|S_i| = (4n-1)p$ for all $i$, and $|S_{i,j}| = r = (2n-1)p$ for all $i, j$. Rules (b) and (c) tell us that for any set of three distinct club members $\{i, j, k\}$, the number of walks featuring $i$ but not $j$ or $k$ is equal to the number of walks featuring $j$ and $k$ but not $i$. We may write this as
$$|S_i| - |S_{i,j}| - |S_{i,k}| + |S_{i,j,k}| = |S_{j,k}| - |S_{i,j,k}|$$
or (rearranging)
$$\begin{aligned} 2|S_{i,j,k}| &= |S_{i,j}| + |S_{i,k}| + |S_{j,k}| - |S_i| \\ &= 3(2n-1)p - (4n-1)p \\ &= 2(n-1)p \end{aligned}$$
or $|S_{i,j,k}| = (n-1)p$. As this formula holds for any set of three distinct club members $\{i, j, k\}$, the result follows.