The answer is $\boxed{C = \frac{3}{2}} $. We split the proof into two parts. Part 1: Proving $C = \frac{3}{2}$ works. For this, we have to show that for any $(a_i)_{i=1}^n $ satisfying the problem conditions, we can choose $b_i$ satisfying the problem conditions. Suppose for some sequence $(a_i)$ we could not choose such $b_i$. In other words, all sequences $b_i\in \{\lfloor a_i \rfloor, \lfloor a_i \rfloor + 1\}$ have their sum of reciprocals either at most $1$, or greater than $C$. For the moment, let $b_i = \lfloor a_i \rfloor + 1$ and every second, randomly subtract some $b_i$ that hasn't been subtracted before by $1$. Each subtraction increases the value of $\sum_{i=1}^n \frac{1}{b_i}$. Hence after $n$ seconds, all $b_i$ are equal to $\lfloor a_i \rfloor$. Since\[ \sum_{i=1}^n \frac{ 1}{ \lfloor a_i \rfloor + 1} < \sum_{i=1} ^n \frac{1}{a_i} = 1 \text{ and } \sum_{i=1}^n \frac{ 1}{ \lfloor a_i \rfloor } > \sum_{i=1}^n \frac{1}{a_i} = 1\]we have that $\sum_{i=1}^n \frac{1}{b_i}$ goes from less than $1$ to more than $1$. Consider the unique second where the sum switches from less than $1$ to more than $1$ (since the sum is strictly increasing such a second must exist). However, when $\sum_{i=1}^n \frac{1}{b_i}$ is greater than $1$, it must also exceed $\frac{3}{2}$ by our assumption. Hence within one second, we go from less than $1$ to more than $\frac{3}{2}$, which means our switch of some $b_k$ increases the sum by more than $\frac 12$, which is impossible, since $\frac{1}{b_k} - \frac{1}{b_{k} + 1} = \frac{1}{b_k (b_k + 1)} \le \frac{1}{2}$. Thus, we have a contradiction, and we can always choose $b_i$ satisfying the problem conditions. Part 2: Proving $C \ge \frac 32$. Suppose there existed a working $C$ less than $\frac 32$. For each $n$, consider the sequence $a_1 = a_2 = \cdots = a_{n-1} = \frac{4n+1}{2}$ and $a_n = \frac{4n + 1}{2n+3}$. We see that $\sum_{i=1}^n \frac{1}{a_i} = 1$, so $(a_i)$ is valid. It's clear that $b_1, b_2 \ldots, b_{n-1}$ are all in the set $\{2n,2n+1\}$ and $b_n \in \{1,2\}$. Consider some $b_i$ sequence where $1 < S = \sum_{i=1}^n \frac{1}{b_i} \le C$. Claim: $b_n = 1$ Proof: Suppose $b_n = 2$ held. Then for each $i < n$, the maximum value for $\frac{1}{b_i}$ is $\frac{1}{2n}$. Hence $S \le \frac 12 + (n-1) \cdot \frac{1}{2n} < 1$, absurd. $\square$ Now, for $i < n$, the minimum value for $\frac{1}{b_i}$ is $\frac{1}{2n+1}$, so $S \ge 1 + \frac{n-1}{2n+1} = \frac{3}{2} - \frac{3}{ 4n + 2}$. Notice that if we set $n$ large, we get that $\frac{3}{4n + 2} < \frac{3}{2} - C$ (for example, any $n > \left(\frac{1}{ 3/2 - C} -2\right)\cdot \frac{1}{4} $ satisfies this), so $S \ge \frac 32 - \frac{3}{4n+2} > C$, absurd. Therefore we must have $C \ge \frac 32$.

The answer is $C=\frac{3}{2}$. Let $S=\sum_{i=1}^n b_i$. For sufficiency, start with all $b_i$ equal to $\lfloor a_i\rfloor$, so that $S > 1$. While $S>\frac{3}{2}$, we can choose some $i$ with $b_i = \lfloor a_i\rfloor$ and set $b_i=\lceil a_i\rceil$. This decreases $S$ by at most $\frac{1}{1}-\frac{1}{2}=\frac{1}{2}$, so $S$ will remain greater than $1$. Since the minimum possibl value of $S$ is $\sum \frac{1}{\lceil a_i\rceil} < 1$, we can continue modifying this way until $S$ is between $1$ and $\frac{3}{2}$, as desired. For the lower bound, let $a_i = 2n-1.5$ for $1\le i\le n-1$, and let $a_{n} = \frac{2n-1.5}{n-0.5}$. Then $b_{n}$ must be $1$ for $S$ to be greater than $1$, so $C\ge 1 + \frac{n-1}{2n-1}$. Letting $n$ be sufficiently large, we see that $C\ge \frac{3}{2}$, as desired.

Sorry but my thread was 2 minutes before yours. @below: in my clock it was marked as 1 pm (which is midnight eastern time converted to my local time).

MathLuis wrote: Sorry but my thread was 2 minutes before yours You posted the thread before 12 PM EST...

rip $\frac{4}{3}$ fakesolve

We present two solutions. The first gives an answer of $C=\tfrac{3}{2}$, and the second gives $C=\tfrac{3n-2}{2n-1}=\tfrac{3}{2}-\tfrac{1}{4n-2}$ for each specific value of $n$, which implies $C=\tfrac{3}{2}$. Let $A_i=\tfrac{1}{a_i}$, $B_i=\tfrac{1}{b_i}$, $\uparrow_i=\tfrac{1}{\lfloor a_i \rfloor}$, and $\downarrow_i=\tfrac{1}{\lceil a_i \rceil}$ for $i=1,\ldots,n$. Both solutions have the same construction: consider \[A_1,A_2,\ldots,A_n=\frac{1}{2}+(n-1)\varepsilon,\frac{1}{2n-2}-\varepsilon,\ldots,\frac{1}{2n-2}-\varepsilon\]for a positive real number $\varepsilon<\tfrac{1}{2n-2}-\tfrac{1}{2n-1}$. If $B_1=\tfrac{1}{2}$, then \[B_1+B_2+\cdots+B_n \le \frac{1}{2}+\frac{1}{2n-2}+\cdots+\frac{1}{2n-2}=1,\]a contradiction. Thus, we must have $B_1=1$, so \[B_1+B_2+\cdots+B_n \ge 1+\frac{1}{2n-1}+\cdots+\frac{1}{2n-1}=\frac{3n-2}{2n-1}.\]By choosing arbitrarily large values of $n$, we obtain $C \ge \tfrac{3}{2}$. Solution 1: We prove that $C=\tfrac{3}{2}$ works. Consider the sums \begin{align*} \uparrow_1+\uparrow_2&+\cdots+\uparrow_n \\ \downarrow_1+\uparrow_2&+\cdots+\uparrow_n \\ \downarrow_1+\downarrow_2&+\cdots+\uparrow_n \\ &\vdotswithin{+} \\ \downarrow_1+\downarrow_2&+\cdots+\downarrow_n. \end{align*}Notice that the first sum is greater than $1$, the last sum is less than $1$, and each sum is less than the previous. Furthermore, the difference between any two consecutive sums is $\uparrow_i-\downarrow_i$ for some $i$, which is at most $\tfrac{1}{2}$. Thus, at least one of these sums lies in $(1,\tfrac{3}{2}]$, as desired. $\square$ Solution 2: We prove that $C=\tfrac{3n-2}{2n-1}$ works for any specific $n$. Assume WLOG that $A_1 \ge \cdots \ge A_n$. If $A_1<\tfrac{1}{2}$, then consider the sums \begin{align*} \uparrow_1+\uparrow_2&+\cdots+\uparrow_n \\ \downarrow_1+\uparrow_2&+\cdots+\uparrow_n \\ \downarrow_1+\downarrow_2&+\cdots+\uparrow_n \\ &\vdotswithin{+} \\ \downarrow_1+\downarrow_2&+\cdots+\downarrow_n. \end{align*}Notice that the first sum is greater than $1$, the last sum is less than $1$, and each sum is less than the previous. Furthermore, the difference between any two consecutive sums is $\uparrow_i-\downarrow_i$ for some $i$, which is at most $\tfrac{1}{6}$. Thus, at least one of these sums lies in $(1,\tfrac{7}{6}] \subset (1,C]$, as desired. Otherwise, we have $A_1 \in (\tfrac{1}{2},1)$ and $A_2<\tfrac{1}{2}$. Consider the sums \begin{align*} s&=\tfrac{1}{2}+\uparrow_2+\cdots+\uparrow_n \\ S&=1+\downarrow_2+\cdots+\downarrow_n. \end{align*} Claim: $s>1$ or $S \le \tfrac{3n-2}{2n-1}$. Proof: Assume FTSOC that both are false. Then, we have $\uparrow_2+\cdots+\uparrow_n \le \tfrac{1}{2}$ and $\downarrow_2+\cdots+\downarrow_n>\tfrac{n-1}{2n-1}$. Notice that $\uparrow_i=\tfrac{1}{1/\downarrow_i-1}=\tfrac{\downarrow_i}{1-\downarrow_i}$. Since $f(x)=\tfrac{x}{1-x}=\tfrac{1}{1-x}-1$ is convex and strictly increasing in $(0,1)$, Jensen gives \[\uparrow_2+\cdots+\uparrow_n=f(\uparrow_2)+\cdots+f(\uparrow_n)>(n-1)f\left(\frac{1}{2n-1}\right)=\frac{1}{2},\]a contradiction. $\square$ If $S \le \tfrac{3n-2}{2n-1}$, then $S \in (1,C]$, so we are done. If $s>1$, then at least one of the sums \begin{align*} s=\tfrac{1}{2}+\uparrow_2&+\cdots+\uparrow_n \\ \tfrac{1}{2}+\uparrow_2&+\cdots+\uparrow_n \\ \tfrac{1}{2}+\downarrow_2&+\cdots+\uparrow_n \\ &\vdotswithin{+} \\ \tfrac{1}{2}+\downarrow_2&+\cdots+\downarrow_n<1 \end{align*}lies in $(1,\tfrac{7}{6}) \subset (1,C]$ using the same argument as the $A_1<\tfrac{1}{2}$ case. $\square$

The answer is $\frac{3}{2}$. (will there even be equality?) We force the answer to be at least $\frac{3}{2}$ by selecting a large $k$, setting $n=k+1$, $a_1 = \frac{2k}{k+0.5}$ and $a_2 = a_3 = \dots = a_{k+1} = \frac{2k^2}{k-0.5}$. Say we round $a_1$ up to $2$. Then, $a_2, a_3, \dots, a_{k+1}$ will round down to $2k+1$, so their reciprocals are at most $\frac{1}{2k+1}$, which is insufficient for $\tfrac{1}{b_1} + \tfrac{1}{b_2} + \dots + \tfrac{1}{b_n} > 1$. Thus we must round $a_1$ down to $1$; then, $a_2, a_3, \dots, a_{k+1}$ round up to $2k+2$, which makes the sum equal to \[ \frac{3k+2}{2k+2},\]arbitrarily close to $\frac{3}{2}$. To attain $\frac{3}{2}$ just start with every number rounded up and pick one number to round down each second. The sum of their reciprocals starts less than $1$, ends greater than $1$ and jumps by at most $\frac{1}{2}$ at each step so we must hit some number in the range $(1, \frac{3}{2} ]$.

I guess this is the official thread now The answer is $C=3/2$. Necessity: Fix $n$, and set $a_2=\cdots=a_n=2n-\tfrac{1}{2}$, and let $a_1=\tfrac{4n-1}{2n+1}$; it's easy to verify that $\tfrac{1}{a_1}+\cdots+\tfrac{1}{a_n}=1$. Clearly we have $b_i \in \{2n-1,2n\}$ for all $2 \leq i \leq n$ and $b_1 \in \{1,2\}$. On the other hand, if $b_1=2$ then $\tfrac{1}{b_1}+\cdots+\tfrac{1}{b_n}<1$, so $b_1=1$ and now $\tfrac{1}{b_1}+\cdots+\tfrac{1}{b_n} \geq 1+\tfrac{n-1}{2n}$. Sending $n \to +\infty$ implies $C \geq 3/2$. $\blacksquare$ Sufficiency: We allow $b_i$ to vary and let $S$ denote the changing value of $\tfrac{1}{b_1}+\cdots+\tfrac{1}{b_n}$ for convenience. We start with each $b_i$ equal to $\lfloor a_i\rfloor+1$; in this case, $S<\tfrac{1}{a_1}+\cdots+\tfrac{1}{a_n}=1$. Now, in increasing order of $1 \leq i \leq n$, switch $b_i$ to $\lfloor a_i\rfloor$. Each time this happens, $S$ increases by $$\frac{1}{\lfloor a_i\rfloor}-\frac{1}{\lfloor a_i\rfloor+1}=\frac{1}{\lfloor a_i\rfloor(\lfloor a_i\rfloor+1)} \leq \frac{1}{2}.$$Furthermore, when this process ends and $b_i=\lfloor a_i\rfloor$ for all $i$ we clearly have $S>\tfrac{1}{a_1}+\cdots+\tfrac{1}{a_n}=1$. Then, if the final value of $S$ is at most $3/2$ we're immediately done. Otherwise, "discrete continuity" guarantees that, at some point, we had $S \in (1,3/2]$, since "crossing" directly from $(0,1]$ to $(3/2,\infty)$ would imply an increase in $S$ of more than $1/2$. $\blacksquare$

Posting here just in case... I claim that $C=\frac{3}{2}$ works, we divide the problem in two claims. Claim 1: $\frac{3}{2} \ge C$ Proof: Suppose FTSOC that $C>\frac{3}{2}$, we will start with all $b_i=\lfloor a_i \rfloor +1$ and then start shifting until we get the sum of $\frac{1}{b_i}$'s to get past 1, $C>\frac{3}{2}$ means that for some $a_1,a_2, ..., a_n$ the moment we get that sum past $1$, then it will also get past $\frac{3}{2}$, for any positive integer $n$ note that $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)} \le \frac{1}{2}$, so in fact this means that adding at most $0.5$ to a thing less than $1$ will make it greater than $1.5$ which is ofc, a contradiction!, thus done. Claim 2: $C \ge \frac{3}{2}$ Proof: Set $a_1=2-\epsilon$ and $a_2=a_3= ...=a_n=\frac{(2-\epsilon)(n-1)}{1-\epsilon}$ for some $\epsilon>0$ small enough. If $b_1=2$ then even if u took the minimal value of the rest of $b_i's$ (which is $2n-2$ by setting $\epsilon$ small enough such that $\frac{1}{\epsilon} \ge n$) we would get $\sum_{i=1}^{n} \frac{1}{b_i}$ to be at most $1$, hence we are forced to have $b_1=1$, and ofc then we have $C \ge 1+\frac{n-1}{2n-1}$ and by setting $n$ large enough we ofc have $C \ge \frac{3}{2}$ as desired thus we are done. Hence we have the minumal $C$ to be $C=\frac{3}{2}$ as desired, thus we are done .

$2-\varepsilon$, $\frac{2-\varepsilon}{1-\varepsilon}(n-1)$ $n-1$ times gives $C\geq\frac32$. WLOG $a_1=\min(a_1,a_2,\ldots,a_n)$. If $a_1<2$, select $b_1=1$ and $b_i=\lfloor a_i\rfloor+1$, giving a bound of $1+1-\frac1{a_1}<\frac32$. Otherwise, select $b_i=\lfloor a_i\rfloor$, so $\frac{b_i}{a_i}<\frac32$. Therefore, $C=\boxed{\frac32}$.

This problem feels like a Putnam-style problem. Construction: (Can someone tell me if this is rigorous enough?) Consider sequence $\frac{1}{2 - x}$, $\frac{1}{n - y}$ repeated c times, and $\frac{1}{z}$ for a sufficiently small x, y and sufficiently large n. such that $z \ge n - y$. Note that $\frac{1}{2 - x}$ must have floor applied to form $b_1$ as otherwise, the total sum . Since n approaches infinity, the fraction $\frac{1}{n - y}$ becomes $\frac{1}{n}$ at the least so will increase by at most $\frac{y}{n(n - y)}$. Since $c < n$, sum of all such fractions can be reduced by at most $\frac{y}{n - y}$. For $\frac{1}{z}$, can be reduced by less than $\frac{1}{n^2}$. Therefore, as x, y approach 0 and n approaches infinity, other fractions besides $\frac{1}{2 - x}$ cannot be reduced by a finite amount so any value below 1 + 1/2 = 3/2 is possible to achieve. Proof of Bound: If all $a_i$ are bigger than 2, then can see that the fraction denominator will increase by at most 1 meaning overall all fractions increase by less than a factor of $\frac32$. If exists an $a_k < 2$, (only 1 $a_i$ can be less than 2), then all remaining fractions can decrease by at most a factor of 2. Let $\frac{1}{a_k} = x$. Floor $a_k$ so that $a_k$ and $x$ become 1. Ceiling everything else so that all other fractions decrease by at most $\frac{1 - x}{2}$. Therefore new sum is between $2 - x$ and $1 + \frac{1 - x}{2}$ which are both $\ge 1$ and $\le \frac32$.

Here is the official solution write-up. Answer. The answer is $C=\frac{3}{2}$. Lower bound. Note that if $a_1 = \frac{4n-3}{2n-1}$ and $a_i = \frac{4n-3}{2}$ for $i > 1$, then we must have $b_1 \in \{1,2\}$ and $b_i \in \{2n-2,2n-1\}$ for $i > 1$. If we take $b_1 = 2$ then we obtain \[ \frac{1}{b_1} + \frac{1}{b_2} + \dots + \frac{1}{b_n} \le \frac{1}{2} + (n-1) \cdot \frac{1}{2n-2} = 1, \]whereas if we take $b_1 = 1$ then we obtain \[ \frac{1}{b_1} + \frac{1}{b_2} + \dots + \frac{1}{b_n} \ge 1 + (n-1) \cdot \frac{1}{2n-1} = \frac{3n-2}{2n-1}. \]This shows that $C \ge \frac{3n-2}{2n-1}$, and as $n\to \infty$ this shows that $C\geq \frac{3}{2}$. Upper bound. For $0\leq k\leq n$, define \[ c_i = \sum_{i=1}^{k}\frac{1}{\lfloor a_i\rfloor}+\sum_{i=k+1}^{n}\frac{1}{\lfloor a_i\rfloor+1}. \]Note that $c_0 < c_1 < \dots < c_n$ and \[c_0 < \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} = 1 < c_n.\]This means there exists a unique value of $k$ for which $c_{k-1} < 1 < c_{k}$. For this $k$ we have \[ 1 < c_{k} = c_{k-1} + \frac{1}{(\lfloor a_k\rfloor)(\lfloor a_k\rfloor+1)} < 1 + \frac{1}{1\cdot 2} = \frac{3}{2}. \]Therefore we may choose $b_i = \lfloor a_i\rfloor$ for $i\leq k$ and $b_i = \lfloor a_i\rfloor+1$ for $i > k$. Upper bound (alternate). First suppose $a_i < 2$ for some $i$. Assume without loss of generality $i=1$ here. Let $b_1=1$ and $b_i=\lfloor a_i\rfloor+1$ for all other $i$. Then \begin{align*} 1 &< \frac{1}{b_1} + \dots + \frac{1}{b_n} = 1 + \frac{1}{\lfloor a_2\rfloor + 1} + \dotsb + \frac{1}{\lfloor a_n\rfloor + 1} \\ &< \left(\frac{1}{2} + \frac{1}{a_1}\right) + \frac{1}{a_2} + \dotsb + \frac{1}{a_n} = \frac{3}{2}. \end{align*}Now suppose $a_i > 2$ always. Then $\frac{a_i}{\lfloor a_i\rfloor} < \frac{3}{2}$, so \[ 1 = \frac{1}{a_1} + \dotsb + \frac{1}{a_n} < \frac{1}{\lfloor a_1\rfloor} + \dots + \frac{1}{\lfloor a_n\rfloor} < \frac{3}{2}\left(\frac{1}{a_1} + \dotsb + \frac{1}{a_n}\right) = \frac{3}{2}. \]Therefore we may let $b_i=\lfloor a_i\rfloor$ for all $i$. Remark: The original proposal asked to find the optimal $C$ for a fixed $n$. The answer is $\frac{3n-2}{2n-1}$, i.e. the lower bound construction in the solution is optimal.

v_Enhance wrote: Here is the official solution write-up. Answer. The answer is $C=\frac{3}{2}$. Lower bound. Note that if $a_1 = \frac{4n-3}{2n-1}$ and $a_i = \frac{4n-3}{2}$ for $i > 1$, then we must have $b_1 \in \{1,2\}$ and $b_i \in \{2n-2,2n-1\}$ for $i > 1$. If we take $b_1 = 2$ then we obtain \[ \frac{1}{b_1} + \frac{1}{b_2} + \dots + \frac{1}{b_n} \le \frac{1}{2} + (n-1) \cdot \frac{1}{2n-2} = 1, \]whereas if we take $b_1 = 1$ then we obtain \[ \frac{1}{b_1} + \frac{1}{b_2} + \dots + \frac{1}{b_n} \ge 1 + (n-1) \cdot \frac{1}{2n-1} = \frac{3n-2}{2n-1}. \]This shows that $C \ge \frac{3n-2}{2n-1}$, and as $n\to \infty$ this shows that $C\geq \frac{3}{2}$. Upper bound. For $0\leq k\leq n$, define \[ c_i = \sum_{i=1}^{k}\frac{1}{\lfloor a_i\rfloor}+\sum_{i=k+1}^{n}\frac{1}{\lfloor a_i\rfloor+1}. \]Note that $c_0 < c_1 < \dots < c_n$ and \[c_0 < \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} = 1 < c_n.\]This means there exists a unique value of $k$ for which $c_{k-1} < 1 < c_{k}$. For this $k$ we have \[ 1 < c_{k} = c_{k-1} + \frac{1}{(\lfloor a_k\rfloor)(\lfloor a_k\rfloor+1)} < 1 + \frac{1}{1\cdot 2} = \frac{3}{2}. \]Therefore we may choose $b_i = \lfloor a_i\rfloor$ for $i\leq k$ and $b_i = \lfloor a_i\rfloor+1$ for $i > k$. Upper bound (alternate). First suppose $a_i < 2$ for some $i$. Assume without loss of generality $i=1$ here. Let $b_1=1$ and $b_i=\lfloor a_i\rfloor+1$ for all other $i$. Then \begin{align*} 1 &< \frac{1}{b_1} + \dots + \frac{1}{b_n} = 1 + \frac{1}{\lfloor a_2\rfloor + 1} + \dotsb + \frac{1}{\lfloor a_n\rfloor + 1} \\ &< \left(\frac{1}{2} + \frac{1}{a_1}\right) + \frac{1}{a_2} + \dotsb + \frac{1}{a_n} = \frac{3}{2}. \end{align*}Now suppose $a_i > 2$ always. Then $\frac{a_i}{\lfloor a_i\rfloor} < \frac{3}{2}$, so \[ 1 = \frac{1}{a_1} + \dotsb + \frac{1}{a_n} < \frac{1}{\lfloor a_1\rfloor} + \dots + \frac{1}{\lfloor a_n\rfloor} < \frac{3}{2}\left(\frac{1}{a_1} + \dotsb + \frac{1}{a_n}\right) = \frac{3}{2}. \]Therefore we may let $b_i=\lfloor a_i\rfloor$ for all $i$. Remark: The original proposal asked to find the optimal $C$ for a fixed $n$. The answer is $\frac{3n-2}{2n-1}$, i.e. the lower bound construction in the solution is optimal. What is the MOHS rating of this problem please ?

We uploaded our solution https://calimath.org/pdf/USATST2024-1.pdf on youtube https://youtu.be/k0_M6o8O8rY.

The answer is $C=\frac32$. All unindexed summations are from $1$ to $n$.

Interesting how I got $C=\frac43$ as the answer due to a calculation mistake, and only found it when I was at my 4th out of 6 cases.

Answer: $\boxed{C=\frac{3}{2}}$ Lower Bound: $$\frac{1}{2-\epsilon}+\underbrace{\frac{1}{2k+\epsilon}+\frac{1}{2k+\epsilon}+\dots+\frac{1}{2k+\epsilon}}_{k\text{ terms}}$$If we round $2-\epsilon$ to $2$ then the sum will be smaller than $1$. So the least the sum could be that is greater than $1$ would be $1+\frac{k}{2k+1}$. Taking the limit as $k\rightarrow \infty$ we get $\frac{3}{2}\leq C$ Upper Bound: Notice that $a_i>1$. To start let all of the $b_i=\lfloor{a_i}\rfloor+1$. Changing one of the $b_i$ to $\lfloor{a_i}\rfloor$ will increase the sum by at most $\frac{1}{2}$ and since if you changed all of them the sum will be greater than one at some point it must lie within the desired range.

The answer is $\frac{3}{2}$. For the construction, let $a_1= 2-\epsilon$ while $a_i=\frac{2-\epsilon}{1-\epsilon}$ for all $i>1$. Now for the bound, if we have that $a_1<2$ then take $b_1=\lfloor{a_1}\rfloor$, and all other $b_i=\lfloor{a_i}\rfloor+1$. Then we have that. \[ 1 < \frac{1}{b_1} + \dots + \frac{1}{b_n} = 1 + \frac{1}{\lfloor a_2\rfloor + 1} + \dotsb + \frac{1}{\lfloor a_n\rfloor + 1} < \frac{3}{2} \] If all $a_i>2$ then take all $b_i=\lfloor{a_i}\rfloor$, \[ 1 = \frac{1}{a_1} + \dotsb + \frac{1}{a_n} < \frac{1}{\lfloor a_1\rfloor} + \dots + \frac{1}{\lfloor a_n\rfloor}\] Which is less than $\frac{3}{2}$, as the ratio of $a_i$ to $\lfloor{a_i}\rfloor$ is less than $3/2$. Hence we are done

unfortunately got spoiled

The answer is $\tfrac{3}{2}$. For brevity, let $S=\tfrac{1}{b_1} + \tfrac{1}{b_2} + \cdots + \tfrac{1}{b_n}$. First we prove that $C\ge \tfrac{3}{2}$. We pick $a_1=2-\tfrac{1}{2n-1}$ and $a_2=a_3=\dots=a_n=2n-\tfrac{3}{2}$. This satisfies the identity because \[\frac{1}{2-\tfrac{1}{2n-1}}+\frac{n-1}{2n-\tfrac{3}{2}}=\frac{2n-1}{4n-3}+\frac{2n-2}{4n-3}=1\]Note that $b_1\in \{1,2\}$ while $b_2,b_3,\dots,b_n\in \{2n-2,2n-1\}$. Claim: $b_1=1$ Suppose $b_1=2$ then \[S\le \frac{1}{2}+(n-1)\cdot \frac{1}{2n-2}=1\]which is a contradiction. Since $b_1=1$, we have \[S\ge 1+(n-1)\cdot \frac{1}{2n-1}\]Taking the limit as $n\to \infty$, we have $C\ge \tfrac{3}{2}$ as desired. Next we prove that $C=\tfrac{3}{2}$ works. Let \[S_i=\sum_{j=1}^{n}\frac{1}{\lfloor a_i + \left<j>i\right> \rfloor}\]where $\left<x\right>$ for a boolean statement $x$ is $1$ if true and $0$ if false. Note that $S_0<S_1<\dots<S_n$ where $S_0<1$ and $S_n>1$. Therefore, there exists a value $k$ such that $S_k<1<S_{k+1}$. Note that \[S_{k+1}-S_k\le \frac12\]Therefore, we have $S_{k+1}\le S_k+\tfrac12<\tfrac{3}{2}$ as desired.

Our answer is $\boxed{\frac 32}$, constructable with \[(2-\epsilon_1, \underbrace{2k+\epsilon_2, \ldots, 2k+\epsilon_2}_{k \text{ values}}) \implies C = 1 + \frac{k}{2k+1}\] and $k \rightarrow \infty$. Now notice that if we set $\{b_i\} = \{\lfloor a_i \rfloor\}$ and try to "jump" our way back, these jumps cannot be more than $\tfrac 11 - \tfrac 12 = \tfrac 12$, so at some point we must land in the interval $(1, \tfrac 32]$. $\blacksquare$