Let $X$ be an integrable real-valued random variable with a strictly positive probability density function $f$ on $\mathbb R$. What is $\mathbb E \left(X \middle \vert \left|X\right| \right)$?
To me, it is obvious that conditionally on $\left|X\right| = x$, $X$ has a discrete distribution with 2 possible values : $-x$ and $x$. As for the probability of each outcome, I have a strong intuition that $\mathbb P \left(X = x \middle \vert \left|X\right| = x \right) = \frac{f(x)}{f(x) + f(-x)}$. After all, that defines a probability and it is worth $\frac{1}{2}$ for symmetric distributions.
However, I have absolutely no idea how to justify it... Is it even correct ?
Your intuition is correct, but you are using $x$ on both sides of your conditioning, which renders your answer ambiguous and confusing. Instead, write $Y:=|X|$ and use a different variable $y$ for the value of $Y$.
Your reasoning is: given $Y=y$, there are two possible values for $X$, namely $y$ and $-y$, with probabilities proportional to $f(y)$ and $f(-y)$ respectively. Since the two probabilities must sum to unity, the probabilities should be $\frac{f(y)}{f(y)+f(-y)}$ and $\frac{f(-y)}{f(y)+f(-y)}$ respectively.
The expected value of $X$ given $Y=y$ is then a straightforward calculation of the expectation of a discrete random variable: $$E(X\mid Y=y) = y\frac{f(y)}{f(y)+f(-y)}+(-y)\frac{f(-y)}{f(y)+f(-y)}=y\frac{f(y) -f(-y)}{f(y)+f(-y)}$$ and therefore your candidate conditional expectation is $$E(X\mid Y)=Y\frac{f(Y) -f(-Y)}{f(Y)+f(-Y)}. $$ Call the RHS $h(Y)$. To prove $h(Y)$ is indeed the conditional expectation of $X$ given $Y$, simply use the definition, which requires you to verify:
Step 1 is immediate. For step 2 it's enough to check this for any $A$ of the form $\{|X|\le c\}$, since such sets generate $\sigma(Y)$. You now have an exercise in integral calculus.