I thought that if you try and calculate the sum $1+2+3+\dots$ in any "reasonable" way, you should get $\frac{-1}{12}$. One example of such a manipulation is in the Mathologer video: https://www.youtube.com/watch?v=jcKRGpMiVTw&list=PL-LSf4Q-q_0s7dF7MHmnEtN5N8YeYNCw1&index=6&t=363s. Another example: https://medium.com/cantors-paradise/the-ramanujan-summation-1-2-3-1-12-a8cc23dea793
This produces the same answer as the analytic continuation of the Reimann Zeta function and many other methods. However, I just found a route that shows it's $\frac{-1}{8}$. Let:
$$s=1+2^2+3^2+\dots$$ $$4s=2^2+2^2 2^2+2^2 3^3+\dots$$
Subtracting,
$$-3s = 1+3^2+5^2+7^2+\dots$$ Shifting one term to the left: $$-3s = 1+3^2+5^2+\dots$$
Subtracting:
$$0=1+2 \times4+2 \times8+2\times12 +\dots=1+8(1+2+3+\dots)$$
$$=> 1+2+3+\dots = \frac{-1}{8}$$
Where did I go wrong where the others didn't?
The error becomes evident when you consider the analogous finite sum.
Let $$S(n) = \sum_{k=1}^n k^2 = 1^2 + 2^2 + \cdots + n^2.$$ Then following your logic, $$4S(n) = \sum_{k=1}^n (2k)^2,$$ and $$-3S(n) = S(n) - 4S(n) = \sum_{k=1}^n k^2 - (2k)^2.$$ When we perform the cancellation of the even terms, the first sum becomes $$\sum_{j=1}^{\lceil n/2 \rceil} (2j-1)^2$$ as you expect, but the number of terms that are cancelled in the second sum is not the entire sum: what is left over is $$\sum_{j=\lfloor n/2 \rfloor+1}^{n} (2j)^2.$$ So $$-3S(n) = \sum_{j=1}^{\lceil n/2 \rceil} (2j-1)^2 - \sum_{j=\lfloor n/2 \rfloor + 1}^n (2j)^2.$$ Then you shift the summation index by $1$ and perform more calculations. For the sake of clarity, assume $n = 2m$ is even, so that we have $$-3S(2m) = \sum_{j=1}^m (2j-1)^2 - \sum_{j=m+1}^{2m} (2j)^2,$$ consequently your shift and subtraction becomes $$\begin{align*} 0 &\overset{?}{=} 3S(2(m+1)) - 1 - 3S(2m) \\ &= \left( \sum_{j=2}^{m+1} (2j-1)^2 - \sum_{j=m+2}^{2m+2} (2j)^2 \right) - \left(\sum_{j=1}^m (2j-1)^2 - \sum_{j=m+1}^{2m} (2j)^2\right) \\ &= \sum_{j=1}^m (2j+1)^2 - (2j-1)^2 \\ & \quad - \left((2(2m+2))^2 + (2(2m+1))^2 - (2(m+1))^2 \right) \\ &= 8\sum_{j=1}^m j - (28m^2 + 40m + 16). \end{align*}$$ And here now is where you see that your reasoning fails, because this remainder term is quadratically increasing in $m$, and is not vanishing as $n \to \infty$.