Ireland

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 $$m\left(\genfrac{}{}{0ex}{}{2n}{2}\right)=r\left(\genfrac{}{}{0ex}{}{4n}{2}\right),$$ 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 $\left(\genfrac{}{}{0ex}{}{2n}{2}\right)$ 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) $$\begin{array}{rl}2|S_{i,j,k}|& =|S_{i,j}|+|S_{i,k}|+|S_{j,k}|-|S_i|\\ & =3(2n-1)p-(4n-1)p\\ & =2(n-1)p\end{array}$$ 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.