Look like a flower but be the serpent underneath

Take a look at the summation: $\sum _{ k=\frac { n(n-1) }{ 2 } +1 }^{ \frac { (n)(n+1) }{ 2 } }{ { a }_{ k } } =\sum _{ k={ T }_{ n - 1 }+1 }^{ { T }_{ n} }{ { a }_{ k } }$ Where ${T}_{n}$ is the ${n}^{th}$ triangular number. this makes it easy to understand what the question is asking.

Tilt the matrix such that $1.1$ becomes the top of the "pyramid" and $2.2$ , $1.2$ becomes the second row of the "pyramid" and so on. The summation is just asking for the sum of the terms of the $n$ row.

It can be easily derived that $S(n)=1.1\sum _{ k=1 }^{ n }{ k } =1.1\times{ T }_{ n }$

Which means $S(n)$ would not be an integer unless ${ T }_{ n }$ is divisible by $10$ .

This means that $n(n+1)$ has to be divisible by $20$

You can further derive that $S(n)$ is an integer when $n$ satisfies any of the bellow conditions:

$n=\left\{ \begin{matrix} 20a+20 \\ 20a+19 \\ 20a+15 \\ 20a+4\end{matrix} \right\}$ $a\ge 0$ Where $a$ is an integer

So, to find the sum of all the possible values to $n$ :

$\left( \sum _{ a=0 }^{ 99 }{ 20a+20 } \right) +\left( \sum _{ a=0 }^{ 99 }{ 20a+19 } \right) +\left( \sum _{ a=0 }^{ 100 }{ 20a+15 } \right) +\left( \sum _{ a=0 }^{ 100 }{ 20a+4 } \right) =\boxed{405819}$