Eccoci.

As we all know the equation $x^2=-1$ has no solutions for real $x$.
If we however introduce the imaginary number $i$ this equation has two solutions: $x=i$ and $x=-i$.
If we go a step further the equation $(x-3)^2=-4$ has two complex solutions: $x=3+2i$ and $x=3-2i$.
$x=3+2i$ and $x=3-2i$ are called each others' complex conjugate.
Numbers of the form $a+bi$ are called complex numbers.
In general $a+bi$ and $a-bi$ are each other's complex conjugate.

A Gaussian Integer is a complex number $a+bi$ such that both $a$ and $b$ are integers.
The regular integers are also Gaussian integers (with $b=0$).
To distinguish them from Gaussian integers with $b \ne 0$ we call such integers "rational integers."
A Gaussian integer $a+bi$ is called a divisor of a rational integer $n$ if the result $\dfrac n {a + bi}$ is also a Gaussian integer.
If for example we divide $5$ by $1+2i$ we can simplify $\dfrac{5}{1 + 2i}$ in the following manner:
Multiply numerator and denominator by the complex conjugate of $1+2i$: $1-2i$.
The result is $\dfrac{5}{1 + 2i} = \dfrac{5}{1 + 2i}\dfrac{1 - 2i}{1 - 2i} = \dfrac{5(1 - 2i)}{1 - (2i)^2} = \dfrac{5(1 - 2i)}{1 - (-4)} = \dfrac{5(1 - 2i)}{5} = 1 - 2i$.
So $1+2i$ is a divisor of $5$.
Note that $1+i$ is not a divisor of $5$ because $\dfrac{5}{1 + i} = \dfrac{5}{2} - \dfrac{5}{2}i$.
Note also that if the Gaussian Integer $(a+bi)$ is a divisor of a rational integer $n$, then its complex conjugate $(a-bi)$ is also a divisor of $n$.

In fact, $5$ has six divisors such that the real part is positive: $\{1, 1 + 2i, 1 - 2i, 2 + i, 2 - i, 5\}$.
The following is a table of all of the divisors for the first five positive rational integers:

For divisors with positive real parts, then, we have: $\sum \limits_{n = 1}^{5} {s(n)} = 35$.

$\sum \limits_{n = 1}^{10^5} {s(n)} = 17924657155$.

What is $\sum \limits_{n = 1}^{10^8} {s(n)}$?