Let $\Omega$ be the halting probability, and let $X$ be the set of rational numbers less than $\Omega$. I think I can reasonably claim that $X$ is a natural set.

Since $\Omega$ is left-c.e., $X$ is $\Sigma^0_1$. Also, $X$ is non-computable since it's Turing equivalent to $\Omega$. However, $X$ cannot be many-one complete for $\Sigma^0_1$ sets. Indeed, no convex set of rationals (or finite union of convex sets of rationals) can be many-one complete for $\Sigma^0_1$ sets.

Suppose $X$ were complete. Note that $\emptyset'$ is *uniformly* many-one complete: from a $\Sigma^0_1$ index for a set, we can compute an index for its reduction to $\emptyset'$. Since $X$ is complete, there is a reduction from $\emptyset'$ to $X$, and so by composing these we see that $X$ is uniformly many-one complete (this argument works for any complete set).

Now, we'll make a $\Sigma^0_1$ set $V$, and by a standard argument with the recursion theorem we can assume we already know an $n$ with $V = W_n$. Using the uniformity, this means we know the index of a computable reduction $f$ from $V$ to $X$.

Begin by computing $f(0), f(1)$ and $f(2)$. These are rational numbers, and so are arranged in some order. Without loss of generality, let's assume that $f(0) \le_\mathbb{Q} f(1) \le_\mathbb{Q} f(2)$. Then enumerate $0$ and $2$ into $V$, but leave $1$ out. By assumption, this means that $f(0)$ and $f(2)$ are in $X$, but $f(1)$ is not. But this contradicts the convexity of $X$.