This problem is polynomially solvable.
Claim: solving the problem on input $a = (a_1, \ldots, a_n)$ and $b = (b_1, \ldots, b_n)$ using a partition $P$ is equivalent to choosing a multiset of numbers $X$ with $|X| = |P|$ such that for $i = 1, \ldots, n$, there exists an $x \in X$ with $a_i \le x$ and $b_i \le 1-x$.
Assuming this claim, solving your problem is equivalent to finding the smallest set $X$ such that for each $i = 1, \ldots, n$, there exists an $x \in X$ with $a_i \le x$ and $b_i \le 1-x$. For any fixed $i$, this condition can be rewritten as $a_i \le x \le 1-b_i$. Thus, the problem is equivalent to finding the smallest set $X$ which includes at least one point from each interval $[a_i, 1-b_i]$.
It is easy to show that the following algorithm finds an optimal solution:
Algorithm: Initialize $X$ to be the empty set. Scan from 0 to 1. When leaving each interval during this scan, check whether the interval you are leaving already contains a point in $X$. If yes, continue without doing anything. If no, add the current value of the scan-line (i.e. the end of the interval) to $X$. Once you reach the end of the scan, output $X$.
Thus, all that's left is to prove the claim:
proof of claim:
First suppose we have a partition $P$ solving your problem. Then define $X = \{x_S~|~S \in P\}$ where $x_S = \max_{i \in S}a_i$.
If $i' \in \{1, \ldots, n\}$, then $i' \in S$ for some $S \in P$.
Since $i' \in S$, clearly we have that $a_{i'} \le max_{i \in S}a_i$. The RHS, however, is the definition of $x_S$, so this shows $a_{i'} \le x_S$.
By the conditions on the partition, $max_{i \in S}a_i + max_{i \in S}b_i \le 1$, or in other words $max_{i \in S}b_i \le 1 - max_{i \in S}a_i = 1 - x_S$. Simply applying the fact that $i' \in S$, we have that $b_{i'} \le max_{i \in S}b_i$. Putting this together, we see that $b_{i'} \le 1-x_S$.
Thus we have shown that for $i = 1, \ldots, n$, there exists an $x \in X$ with $a_i \le x$ and $b_i \le 1-x$.
Next suppose that we have a set $X$ such that for $i = 1, \ldots, n$, there exists an $x \in X$ with $a_i \le x$ and $b_i \le 1-x$.
Name the elements of $X$ as $x_1, x_2, \ldots, x_{|X|}$. Then let $S_j = \{i \in \{1,2,\ldots,n\}~|~a_i \le x_j ~\text{and}~ b_i \le 1-x_j\}$. By the property of $X$, every element of $\{1,2,\ldots,n\}$ is in at least one $S_j$. Then if we define $S_j' = \{i \in S_j~|~i \not\in S_{j'} ~\text{for}~j'<j\}$, we see that sets $S_1', \ldots, S_{|X|}'$ form a partition of $\{1,2,\ldots,n\}$.
We claim that this partition is a valid solution to your problem. Consider any part $S_j'$ in this partition. Since $S_j' \subseteq S_j$, we have that $\max_{i \in S_j'}a_i \le \max_{i \in S_j}a_i \le x_j$
and $\max_{i \in S_j'}b_i \le \max_{i \in S_j}b_i \le 1-x_j$,
and so we see that $\max_{i \in S_j'}a_i + \max_{i \in S_j'}b_i \le x_j + 1-x_j = 1$. This is sufficient to show that the partition is a valid solution to your problem.
