Divisible by sum of naturals

The left- hand side of the above equations can be written as, $\frac{12n}{n(n+1)}$ = $\frac{12}{n+1}$ Since this yields a positive integer, $n+1$ must be a factor of $12$

So, $(n+1)\in$ { $1,2,3.4.6,12$ } or $n\in$ { $1,2,3,5,11$ }

Therefore, $S$ = $1+2+3+5+11$ = $\boxed {22}$

$\frac {6n}{1+2+3+....n}$ =k

$\Rightarrow$ 6n $\frac {2}{n(n+1)}$ =k

$\Rightarrow$ $\frac {12}{n+1}$ =k

Since by given, n & k are natural numbers $\Rightarrow$ n+1 is a factors of 12

$\Rightarrow$ n+1=2; 3; 4; 6; 12

$\Rightarrow$ n=1; 2; 3; 5; 11

$\Rightarrow$ sum of all values of n is 1+2+3+5+11=22

Therefore sum of all values of n is $\boxed{22}$

1+2+3+..+n=(n/2)(1+n).. so 6n/[(n/2)(1+n)]=k 12/(1+n)=k 1+n=1,2,3,4,6,12 in order for k to be a natural number so n=0,1,2,3,5,11 and sum of all such n=22

1 + 2 + 3 + ................ + n = n(n + 1)/2

Then

12/(1 + n) = k

Since k is a natural number

Then (1 + n) is a factor of 12

So

n = 1 , 2 , 3 , 5 , 11

The sum of n's = 22

The LHS of the eq. after simplifying is 12/(n+1)=k.(using,1+2+3+...+n=n(n+1)/2) since K is positive integer, (n+1) is a factor of 12. therefore, n+1+{1,2,3,4,6,12} . So, n={1,2,3,5,11} So , summation of all n =1+2+3+5+11=22