Is the decision problem, for a Turing Machine are there any input strings rejected decidable?
Let us start by properly defining your language as say
$$L = \{\langle M \rangle \mid M \text{ rejects some input } x \in \Sigma^\ast\}$$
where $\langle \cdot \rangle$ denotes some suitable encoding over our alphabet $\Sigma$.
To reduce $L$ to the halting problem $\mathrm{Halt} = \{\langle M, w \rangle \mid M \text{ halts on } w\}$ we consider some $\mathrm{Halt}$ instance consisting of a TM $M$ and some input $w$ and define a new TM $M_w$ which always, no matter what input it is given, just simulates the behavior of $M$ on $w$ with the slight change that it rejects if $M$ accepts, i.e. $M_w$ rejects precisely if $M$ halts on $w$ and does not halt if $M$ does not halt on $w$.
Note that $M_w$ can be algorithmically constructed from $M$ and $w$.
We see that $\langle M, w \rangle \in \mathrm{Halt}$ if and only if $\langle M_w \rangle \in L$, i.e. we have shown that the halting problem reduces to $L$ and thus $L$ is undecidable.
However, $L$ is also recursively enumerable: Fix some ordering of $\Sigma^\ast$ and iteratively execute $M$ on the first $n$ strings of $\Sigma^\ast$ for $n$ steps.
If $M$ rejects some string in $\Sigma^\ast$, this procedure will detect it in finitely many steps, showing that $L$ is semi-decidable and hence recursively enumerable.