Factoraddition

$10^5=(2\times5)^5=2^5\times5 ^5$

So, including $1$ & $10^5$ , divisors of $10^5$ will be,

$2^05^0,2^05^1,2^05^2,2^05^3,2^05^4,2^05^5,2^15^0,......,2^55^3,2^55^4,2^55^5$

$\color{#0C6AC7} {\text{That's where we get the number of total divisors.}}$ $\color{#0C6AC7} {\text{In this case it's, (5+1)(5+1)=36}}$

Thus, sum of the divisors will be, $2^0(5^0+5^1+5^2+5^3+5^4+5^5)+......+2^5(5^0+5^1+5^2+5^3+5^4+5^5)$ $=(2^0+2^1+2^2+2^3+2^4+2^5)(5^0+5^1+5^2+5^3+5^4+5^5)$ $=\left(2^0\times\frac{2^{5+1}-1}{2-1}\right)\left(5^0\times\frac{5^{5+1}-1}{5-1}\right)$ $\color{#0C6AC7} {[\text{As sum of geometric progression till } n^{th} \text{ term is,} \, a\frac{r^{n+1}-1}{r-1}\, \text{where} \, r>1]}$ $=63\times3906$ $=\color{#69047E} {\boxed{246078}}$

The formula for the sum of the divisors of an integer $N$ is

$\displaystyle\prod_{i=1}^{k} \dfrac{p_{i}^{m_{i} + 1} - 1}{p_{i} - 1},$

where the $p_{i}$ 's are the distinct prime factors in the factorization of $N$ and the $m_{i}$ 's are the corresponding multiplicities.

Now $10^{5} = 2^{5}5^{5},$ so we have $p_{1} = 2, m_{1} = 5$ and $p_{2} = 5, m_{2} = 5.$ Plugging these values into the formula gives us that the sum of all positive divisors of $10^{5}$ is

$\dfrac{2^{6} - 1}{2 - 1} \times \dfrac{5^{6} - 1}{5 - 1} = \boxed{246078}.$

Brute force and ignorance for me - using python :

n = 10**5
s = 0
for i in range(1,n+1):
    s += i if n % i == 0 else 0
print s