<s>[INST] Pairs of cards from 1,2,...,n arranged according to rules on pairwise separation - is it possible for various n? 
A deck of $2n+1$ cards consists of a joker and, for each number between $1$ and $n$ inclusive, two cards marked with that number. The $2n + 1$ cards are placed in a row, with the joker in the middle. For each $k$ with $1\le  k\le n$, the two cards numbered $k$ have exactly $k−1$ cards between them. Determine all the values of $n$ not exceeding 10 for which this arrangement is possible. For which values of $n$ is it impossible?

The problem is from the 1992 Canada National Olympiad. I can show by parity arguments that the arrangement is impossible for some $n$. This approach is inconclusive for other values, though.

Numbering positions from $0$ at the left, the joker occupies position $n$ and the rightmost card occupies position $2n$.
Then the sum of positions is given by:
$$S_n = -n + \sum_{k=0}^{2n}{k} = -n + \frac{2n(2n+1)}{2} = 2n^2 \tag{1}$$
Now let the leftmost card of each pair of $k$ be at position $\alpha_k$. Then the rightmost card in the pair is at $\alpha_k + k$, so we have:
$$S_n = \sum_{k=1}^{n}{(2\alpha_k + k)} = \frac{n(n+1)}{2}+2\sum_{k=1}^{n}{\alpha_k}$$ 
Letting the sum of $\alpha$ values be $A$, then we have:
$$S_n = 2A + \frac{n(n+1)}{2} \tag{2}$$
By (1), $S_n$ is even for all $n$, and by (2) $S_n$ is even only if and only if  the second term is even, so we deduce that the arrangement of cards is only possible if $\frac{n(n+1)}{2}$ is even.
This immediately rules out $n=1,2,5,6,9,10$.
For $n=3$, a possible arrangement is:
$$\begin{array}{ccccccc} 1&1&3&\bigstar&2&3&2 \end{array}$$
while for $n=4$, a possible arrangement is:
$$\begin{array}{ccc} 1&1&3&4&\bigstar&3&2&4&2 \end{array}$$
I am not sure how to resolve the question of whether there is a viable arrangement for $n=7$ or for $n=8$.
 [/INST] Possible arrangement for $n=7$:$$\begin{array}{ccc} 1&1&2&5&2&6&7&\bigstar&5&3&4&6&3&7&4 \end{array}$$
Possible arrangement for $n=8$:$$\begin{array}{ccc} 1&1&2&4&2&7&8&4&\bigstar&6&3&5&7&3&8&6&5 \end{array}$$
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