Find the smallest constant $C > 1$ such that the following statement holds: for every integer $n \geq 2$ and sequence of non-integer positive real numbers $a_1, a_2, \dots, a_n$ satisfying $$\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} = 1,$$ it's possible to choose positive integers $b_i$ such that
(i) for each $i = 1, 2, \dots, n$, either $b_i = \lfloor a_i \rfloor$ or  $b_i = \lfloor a_i \rfloor + 1$, and
(ii) we have $$1 < \frac{1}{b_1} + \frac{1}{b_2} + \cdots + \frac{1}{b_n} \leq C.$$
(Here $\lfloor \bullet \rfloor$ denotes the floor function, as usual.)

[i]Merlijn Staps[/i]
1. **Part 1: Proving \( C = \frac{3}{2} \) works.**

   We need to show that for any sequence \( (a_i)_{i=1}^n \) 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 \( b_i = \lfloor a_i \rfloor + 1 \) for all \( i \). Each time we subtract 1 from some \( b_i \), the value of \( \sum_{i=1}^n \frac{1}{b_i} \) increases. Hence, after \( n \) subtractions, all \( b_i \) are equal to \( \lfloor a_i \rfloor \).

   Since
   \[
   \sum_{i=1}^n \frac{1}{\lfloor a_i \rfloor + 1} < \sum_{i=1}^n \frac{1}{a_i} = 1 \quad \text{and} \quad \sum_{i=1}^n \frac{1}{\lfloor a_i \rfloor} > \sum_{i=1}^n \frac{1}{a_i} = 1,
   \]
   we have that \( \sum_{i=1}^n \frac{1}{b_i} \) goes from less than 1 to more than 1. Consider the unique moment where the sum switches from less than 1 to more than 1 (since the sum is strictly increasing, such a moment 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 subtraction, 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 \( \frac{1}{2} \), 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.

2. **Part 2: Proving \( C \ge \frac{3}{2} \).**

   Suppose there existed a working \( C \) less than \( \frac{3}{2} \). 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,
   \[
   S \le \frac{1}{2} + (n-1) \cdot \frac{1}{2n} < 1,
   \]
   which is absurd. \( \blacksquare \)

   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
   \[
   S \ge \frac{3}{2} - \frac{3}{4n+2} > C,
   \]
   which is absurd.

   Therefore, we must have \( C \ge \frac{3}{2} \).

The final answer is \( \boxed{C = \frac{3}{2}} \)