Progression sum

Well obviously we can just do trial and error to get the answer, but where's the fun in that?

Here's how I would have done it:

Note that we can combine the sum of fractions to get $\dfrac{1+2+3+ \cdots + n}{14}$ .

We can state $1 + 2 + \cdots + n$ as the famous identity, $1 + 2 +3 + \cdots + n = \dfrac{n(n+1)}2$ .

So this means that we want to find the second smallest positive integer $n$ such thsat $\dfrac{n(n+1)/2}{14} = \dfrac{n(n+1)}{28}$ is an integer. Or equivalently, $n(n+1)$ must be divisible by 28.

Since $28 = 4\times7$ , then one of $n$ or $n+1$ must be a multiple of 7.

So we are left to check whether one of the solutions below shows that $n(n+1)$ is divisible by 28.

$n = 7\Rightarrow n(n+1) = 56$ is divisible by 28 (this is given to the smallest solution).
$n +1= 7\Rightarrow n(n+1) = 42$ is NOT divisible by 28.
$n = 7\times2 \Rightarrow n(n+1) = 210$ is NOT divisible by 28.
$n +1= 7\times2 \Rightarrow n(n+1) = 182$ is NOT divisible by 28.
$n= 7\times3\Rightarrow n(n+1) = 462$ is NOT divisible by 28.
$n+1= 7\times3\Rightarrow n(n+1) = 420$ is divisible by 28.

Thus, our answer is $7\times3 - 1 = \boxed{20}$ .