The crux of the halting problem is that there can be no Turing machine $M$ such that $\text{Halt}(M(N))=\neg \text{Halt}(N(N))$ for all Turing machines $N$, since $\text{Halt}(M(M))=\neg \text{Halt}(M(M))$ is impossible.
Suppose that we removed the negation, and asked simply for a Turing machine $M$ such that such that $\text{Halt}(M(N))=\text{Halt}(N(N))$ for all $N$. This removes the contradiction. In fact, there is an easy example of such a machine. Namely, we can define $M^{\text{triv}}$ to be the machine which runs $N$ on itself whenever given $N$ as input.
We can now ask: what is $\text{Halt}(M(M))$? A-priori from the relation $\text{Halt}(M(N))=\text{Halt}(N(N))$ we get no clues, since plugging in $N=M$ we get the tautology $\text{Halt}(M(M))=\text{Halt}(M(M))$.
For the trivial machine $M^{\text{triv}}$, we see that $\text{Halt}(M^{\text{triv}}(M^{\text{triv}}))=\text{False}$. This is because when you run $M^{\text{triv}}$ on itself it will recursively run on itself forever, never stopping. In fact, for any naive machine $M$ with $\text{Halt}(M(N))=\text{Halt}(N(N))$ it feels natural that $\text{Halt}(M(M))=\text{False}$. When writing out the definition of $M$ you will have to do some amount of recursion that picks up information about how $N$ runs on itself, so when you run it on itself it will keep on doing this recursion forever. So, this brings us to the question:
Can there be a Turing machine $M$ such that $\text{Halt}(M(N))=\text{Halt}(N(N))$ for all Turing machines $N$, and $\text{Halt}(M(M))=\text{True}$?
One trivial example might be a machine which first detects if its input is equal to itself, halts if that is the case, and runs the input on itself otherwise. However, I am a bit skeptical of this. Can a machine really "know" what its original state was? In trying to determine its original state the Turing machine's state will change. I am by no means an expert in this area, so I can't tell. I have very little intuition about whether or not the central question of my post is non-trivial, or what the answer should be if so, but I am very curious to know the answer.
