Two rows of bounded numbers

The answer is Yes, and it follows from the following claim:

Claim. There exist two non-empty subsets $A\subsetneq \{a_1,\dots a_n\}$ and $B\subsetneq \{b_1,\dots b_n\}$ such that $|sum(A)-sum(B)|\leq 0.5$.

Indeed, replacing the set $A$ with $B$ in the sum $a_1+\dots+a_n$ results in the sum in the interval $[n-0.5,n+0.5]$.

Proof of claim. To prove the claim let's assume that $a_1 \leq a_2\leq \dots \leq a_n$ and $b_1 \leq b_2\leq \dots \leq b_n$. Then it's easy to see that $a_1\leq 1$ and $b_1\leq 1$. If they both do not exceed $0.5$ or both are above $0.5$, the claim holds for $A=\{a_1\}$ and $B=\{b_1\}$.

Without loss of generality it remains to consider the case $b_1\leq 0.5 < a_1$. Clearly, we have $1<b_n\leq n-0.5$. The following lemma implies that we can take $B=\{b_n\}$ and some $A$ whose sum is within distance $0.5$ from $b_n$. QED

Lemma. For any $n\geq 2$, and real numbers $\frac{1}{2}\leq a_1 \leq a_2\leq \dots \leq a_n$ with $a_1+\dots +a_n=n$, the sums of proper subsets of $\{a_1,\dots a_n\}$ form a $0.5$-covering of the interval $[0.5,n-0.5]$.

Proof of lemma. Proof is done by induction on $n$. The statement is trivial for $n=2$. Suppose that the statement holds for all $n\leq m$. Let us prove it for $n=m+1$.

So, for $\frac{1}{2}\leq a_1 \leq a_2\leq \dots \leq a_{m+1}$ with $a_1+\dots +a_{m+1}=m+1$, we need to show that any $t\in[0.5,m+0.5]$ is within distance $0.5$ from the sum of some proper subset of $\{a_1,\dots,a_{m+1}\}$. If $t>\frac{m+1}2$, we replace it with $m+1-t$ (covered by the complement of what covers $t$), and so we assume that $t\leq\frac{m+1}{2}$.

Clearly, we have $a_{m+1}>1$, and thus $$s:=\sum_{i=1}^m a_i = m+1-a_{m+1} < m.$$ We also have $s\geq a_1m \geq \frac{m}{2}$.

Define $t':=t\frac{m}{s}$ and notice that $0.5\leq t<t'\leq \frac{(m+1)m}{2s}\leq m+\frac{m}{2s}$. For $i=1,2,\dots,m$, define $a'_i:=a_i\frac{m}{s}$, which sum to $m$ and have $a'_1\geq a_1\geq \frac12$.

Consider two cases:

• Case 1. If $t'\leq m-0.5$, then by induction, there exists a non-empty subset $J\subsetneq\{1,2,\dots,m\}$ such that $$|\sum_{j\in J} a'_j - t'|\leq 0.5.$$ Multiplying by $\frac{s}{m}$, we get $$|\sum_{j\in J} a_j - t|\leq 0.5\frac{s}{m} \leq 0.5.$$

• Case 2. If $m-0.5 < t' \leq m+\frac{m}{2s}$, we set $J:=\{1,2,\dots,m\}$ giving the same inequality as above.

QED