Yes, one can.
Without loss of generality, assume $a_i\gt0$ and let $I=\mathbb N$. For any subset $A\subseteq\mathbb N$, call
$ W_A=\frac{\sum_{i\in A}a_iu_i}{\sum_{i\in A}a_i} $
the weighted average for $A$. If the $u_i$ all lie in some proper affine subspace of $\mathbb R^n$, we can restrict to that subspace, so without loss of generality we can assume that their affine span is $\mathbb R^n$. Then at some finite index $k$ there are positive contributions from $n+1$ affinely independent vectors $A=\{u_{i_1},\dotsc,u_{i_{n+1}}\}$, and $W_A$ is therefore in the interior of their convex hull $H$. Thus, since both $\sum_ia_i$ and $\sum_ia_iu_i$ converge (so their tails converge to zero) there is some $l\ge k$ such that the weighted average remains inside $H$ if the tail $\{i\in\mathbb N\mid i\ge l\}$ is added to $A$. Then the desired finite sum can be obtained by replacing the contributions from the tail by corresponding linear combinations of $u_{i_1},\dotsc,u_{i_{n+1}}$.