The Eisenstein series $\mathbb{G}_2$ is given by $$\mathbb{G}_2(z) = -\frac{1}{24} + \sum_{n=1}^\infty \sigma_1(n) q^n$$ with $q=e^{2\pi i z}$ and $$\sigma_1(n):=\sum_{d\mid n} d$$ for $n\in\mathbb N$. That's why some authors define $\sigma_1(0):=-\frac{1}{24}$, since then $\mathbb{G}_2(z)$ reads as $$\mathbb{G}_2(z) = \sum_{n=0}^\infty \sigma_1(n) q^n.$$
As you may already know the sum of all natural numbers is $-\frac{1}{12}$.
If we apply our definition of $\sigma_1(n)$ to $n=0$ we get $$\sigma_1(0)=\sum_{d\mid 0} d = \sum_{d=1}^\infty d=-\frac{1}{12}.$$ So in this case the definition of $\sum_{d=1}^\infty d=-\frac{1}{12}$ is inappropriate by a factor of two (we'd rather have $-\frac{1}{24}$ here).
In math I'm used to the principle that everything goes well together. Here it doesn't. Do you have explanations for that? Can this issue be fixed somehow?
Here's the result I came up with. We re-define $\sigma_1(n)$ as: $$\sigma_1(n) := \frac{1}{2}\sum_{d|n}d+\frac{n}{d}$$ Note that this definition preserves the original function since $\frac{n}{d}$ is another divisor and we compensate for adding each divisor twice by dividing the result by 2. Now, evaluating for 0 gives: $$\sigma_1(0) = \frac{1}{2}\sum_{d|0} d+\frac{0}{d} = \frac{1}{2}\sum_{d=1}^{\infty}d = -\frac{1}{24}$$ For being in a situation where rigour is practically impossible, I think this works out alright and it does offer pleasant results. I hope you find it satisfactory.