\begin{align*}\frac11+\frac12+\frac13 +\frac14+ \frac16+\frac{1}{12}\\=\frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12} + \frac{2}{12} +\frac{1}{12} \\=\frac{12+6+4+3+2+1}{12}=\frac{28}{12} = \boxed{\frac{7}{3}}.\end{align*}
The answer is the sum of the divisors of 12 divided by 12, why is that?
Of course it's true "for all values of $12$".
The point is this. The sum of $1/d$ for all divisors of $n$ is the sum of $(n/d)/n$. But the numbers $n/d$ are again the divisors of $n$, i.e. the positive integer $d$ divides $n$ if and only
if there is a positive integer $y$ such that $dy=n$, and then $y=n/d$ is again a divisor of $n$. So
$$ \sum_{d \mid n} \frac{1}{d} = \frac{1}{n} \sum_{d \mid n} d $$