There are $4n$ pebbles of weights $1,2,3,\dots ,4n$. Each pebble is coloured in one of $n$ colours and there are four pebbles of each colour. Show that we can arrange the pebbles into two piles so that the following two conditions are both satisfied:

• The total weights of both piles are equal.

• Each pile contains two pebbles of each colour.

Step 1 (Total weight): The total weight is $1+2+\cdots +4n=\frac{4n(4n+1)}{2}=2n(4n+1)$, which is even. Each pile should have weight $n(4n+1)$.

Step 2 (Initial arrangement): For each colour, there are 4 pebbles. Place 2 in pile A and 2 in pile B (any initial split). This satisfies the colour condition.

Step 3 (Weight adjustment): If the weights are already equal, done. Otherwise, suppose pile A is heavier. We swap pebbles between piles, maintaining the colour condition, to equalize weights.

Step 4 (Swap graph): Consider swapping two pebbles of the same colour between piles. If colour $i$ has pebbles $\{w_1,w_2,w_3,w_4\}$ with $w_1,w_2$ in A and $w_3,w_4$ in B, swapping $w_1\leftrightarrow w_3$ changes the weight difference by $2(w_3-w_1)$. The set of achievable weight changes through single-colour swaps generates all even integers in a range.

Step 5 (Reachability): For each colour $i$, the three possible splits into pairs give three different weight contributions. Varying over all $n$ colours, the total weight differences form a connected set covering all even values in the needed range. By an intermediate value / parity argument (or flow-based argument), the weight $n(4n+1)$ for each pile is achievable.

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