Rounding Percents: how far from 100?

For floors, the optimal bound is $n-1$.

To see why it is a bound, notice that $a_i \leq 1+\lfloor a_i \rfloor$ for every $i$, so that $\sum_{i=1}^{n-1} a_i \leq n-1+\sum_{i=1}^{n-1}\lfloor a_i \rfloor$, and hence $a_n=100-\bigg(\sum_{i=1}^{n-1} a_i\bigg) \geq B$ where $B=100-(n-1+\sum_{i=1}^{n-1}\lfloor a_i \rfloor)$. Now $B$ is an integer, so $a_n\geq B$ implies $\lfloor a_n \rfloor \geq B$, which means that $\sum_{i=1}^{n-1}\lfloor a_i \rfloor \geq 100-(n-1)$. On the other side, we clearly have $\sum_{i=1}^{n}\lfloor a_i \rfloor \leq \sum_{i=1}^{n} a_i=100$. This proves the bound.

To see why the bound is optimal, let $\varepsilon\in(0,\frac{1}{n-1})$ and $a_i=1-\varepsilon$ for $i<n$, and $a_n=(100-(n-1))+(n-1)\varepsilon$.