Certainly! Imagine you have $K=2^k$ copies of the search oracle $U_S$ that you can use. Normally, you'd search by iterating the action
$$
H^{\otimes n}(\mathbb{I}_n-2|0\rangle\langle 0|^{\otimes n})H^{\otimes n}U_S,
$$
starting from an initial state $(H|0\rangle)^{\otimes n}$. This takes time $\Theta(\sqrt{N})$. (I'm using $\mathbb{I}_n$ to denote the $2^n\times 2^n$ identity matrix.)
You could replace this with $2^k$ parallel copies, each indexed by an $x\in\{0,1\}^k$, using
$$
\left(\mathbb{I}_k\otimes H^{\otimes (n-k)}\right)\mathbb{I}_k\otimes(\mathbb{I}_{n-k}-2|0\rangle\langle 0|^{\otimes (n-k)})\left(\mathbb{I}_k\otimes H^{\otimes (n-k)}\right)U_S
$$
and starting from a state $|x\rangle(H|0\rangle)^{\otimes(n-k)}$
The time required for running these would be reduced to $O(\sqrt{N/K})$, at the cost of requiring $K$ times more space.
In a scaling sense, one might consider this an irrelevant result. If you have a fixed number of oracles, $K$, then you get a fixed ($\sqrt{K}$) improvement (just like, if you have $K$ parallel classical cores, the best improvement you can get is a factor of $K$), and that does not change the scaling. But it does change the fundamental running time. We know that Grover's algorithm is exactly optimal. It takes the absolute minimum time possible with a single oracle. So, knowing that you get a $\sqrt{K}$ improvement in time is useful with regards to that benchmark of a specific running time at a specific value of $N$.