Magic Pyramid

Another way to get to $n=4$ directly would be to count in how many lines each circle appears. We see that every circle is in 2 lines except for the top circle which appears in 3 lines. That means the sum of all the circles of all the lines respectively is $2×28 + n = 56+n$ with $n$ as the number on the top. This number must be divisibe by 5 as there are 5 lines. Hence $n$ can only be 4.

The sum of all the numbers from $1$ to $7$ is $28$ . If $n$ is the number at the top, then $28-n$ must divide by $3$ , since the the remaining two circles of the three lines joined by the top circle must add up to the same numbers.

This implies that $n = 1$ , $4$ or $7$ .

Also $28-n$ must be even, since the two horizontal numbers must add up to the same number.

That leaves, $n=\boxed{4}$

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It remains to show that we can achieve the conditions of the problem with $n = 4$ . Indeed we can:

$\begin{array} { c c c } &4& \\ 2&3&7\\ 6&5&1\\ \end{array}$