On the reiteration (or repetition) rule in natural deduction: is the rule "obvious" or does it need a proof?

Yes, the so-called Weakening rule:

if $\Gamma \vdash B$, then $\Gamma, A \vdash B$.

From an "Hilbert-style" point of view, the relevant tautology is:

$\vDash A \to (B \to A)$.

The general meta-logical principle that justifies Reiteration is:

If $\Gamma_1 \vDash \psi$, then $\Gamma_1 \cup \Gamma_2 \vDash \psi$

That is: if $\psi$ is a logical consequence of a set of statements (assumptions) $\Gamma_1$, then $\psi$ is also a logical consequence of those very assumptions $\Gamma_1$ to which we added a further set of assumptions $\Gamma_2$

We can prove this using formal semantics. The general definition of $\Gamma \vDash \psi$ is that there is no truth-assignment (valuation) $h$ such that $h(\phi)=True$ for all $\phi \in \Gamma$ and $h(\psi)=False$.

So, take any $\Gamma_1$, $\Gamma_2$, and $\psi$. Assume $\Gamma_1 \vDash \psi$. By definition, that means that there is no truth-assignment (valuation) $h$ such that $h(\phi)=True$ for all $\phi \in \Gamma_1$ and $h(\psi)=False$. Now let's do a proof by Contradiciton, and let's assume that $\Gamma_1 \cup \Gamma_2 \not \vDash \psi$. That means that there is some $h$ such that $h(\phi)=True$ for all $\phi \in \Gamma_1 \cup \Gamma_2 $ and $h(\psi)=False$. But if $h(\phi)=True$ for all $\phi \in \Gamma_1 \cup \Gamma_2 $, then it is certainly true that $h(\phi)=True$ for all $\phi \in \Gamma_1$. Hence, we have that there is some $h$ for which $h(\phi)=True$ for all $\phi \in \Gamma_1$ and $h(\psi)=False$. This contradicts the earlier claim that there was no such $h$. So, it follows that $\Gamma_1 \cup \Gamma_2 \vDash \psi$. Hence, we have proven our desired result: If $\Gamma_1 \vDash \psi$, then $\Gamma_1 \cup \Gamma_2 \vDash \psi$

There are a lot of slightly different versions of natural deduction.

In many of them, this principle is included as a rule or axiom — in that case, it is usually known as “weakening”, as noted in Mauro Allegranza’s answer.

In some variants, it’s not included as an axiom or rule. So in these versions, if you want to know or use this fact, then yes, it requires proof. As with anything in maths, being “obvious” is never a substitute for proof — it just means that (hopefully) the proof should be straightforward. One such proof (assuming that the completeness theorem has already been shown) is given in Bram28’s answer. Other more elementary proofs are usually possible, working just with derivations and not involving models; but these will unavoidably depend more on which specific version of natural deduction you’re using.