If this is correct, then the demonstration below must have a fault, but I can't find it.
Assuming $1+2+3+\cdots = -\frac{1}{12}$ (1) is true.
Adding $0$ on both sides: $0 + 1+2+3+\cdots = 0 -\frac{1}{12}$ (2)
Subtracting (2) from (1),
we get, $1+1+1+1+1+1+\cdots=0$ (3)
Adding $0$ on both sides again, $0+1+1+1+1+1+\cdots = 0 + 0$ (4)
Now subtracting (4) from (3)
we get, $1=0$, which is a contradiction.
Obviously our assumption that $1+2+3+4+\cdots = -\frac{1}{12}$ is wrong.
There is some error in the steps above, but I cannot find it. What is it?
The error is that you assumed a result that is false when using standard definitions of summation and convergence.
The result $1+2+3+\cdots = -\frac{1}{12} $ is only true when nonstandard methods of summation are allowed.
This is like claiming that $1+2+4+8+... =-1 $ since $\dfrac1{1-x} =1+x+x^2+... $ and substituting $x=2$ to gives $1+2+4+... =\dfrac1{1-2} =-1 $.