Problem 1933

*Proof* 1. Nonexistence. Assume that there exists such a positive integer $n$. Let $x$ be an irrational number. Since the terms of the sequence $\{x\}$, $\{2x\}$, $\dots$ are distinct and densely distributed in the interval $(0, 1)$, there exist infinitely many positive integers $d$ satisfying
$$\{dx\} = \min \{ \{ax\} \mid a = 1, 2, \dots, d \}.$$
Arrange these $d$ in an infinite increasing sequence $d_1 < d_2 < \dots$.
We will now show that for all $n = d_i - 1$ ($i \ge 2$), we always have
$$\min \{ \{kx\} \mid k = 1, \dots, n \} = \{d_{i-1}x\} \ge \frac{1}{d_i}.$$
Suppose that this inequality does not hold, then $d_i \cdot \{d_{i-1}x\} \le 1$. Thus,
$$\{d_{i-1}d_i x\} = \{d_i \cdot \lfloor d_{i-1}x \rfloor + d_i \cdot \{d_{i-1}x\}\} = d_i \cdot \{d_{i-1}x\}.$$
On the other hand,
$$\{d_i d_{i-1} x\} = \{d_{i-1} \lfloor d_i x \rfloor + d_{i-1} \{d_i x\}\} \le d_{i-1} \{d_i x\}.$$
Combining the above two equations, we obtain $d_{i-1} \cdot \{d_i x\} \ge d_i \cdot \{d_{i-1} x\}$. However, $d_{i-1} < d_i$ and $\{d_i x\} < \{d_{i-1} x\}$, which leads to a contradiction!
Therefore, $n = d_i - 1$ satisfies the given condition, and there are infinitely many such $n$. $\square$

*Proof* 2. Nonexistent. If there is no irrational number $x$ satisfying the condition, then there exists $n_0 \in \mathbb{Z}_+$ such that for any $n \ge n_0$, there exists $k \in \{1, \dots, n\}$ such that $\{kx\} < \frac{1}{n+1}$. Since $x$ is an irrational number, for any $k \in \mathbb{Z}_+$, $\{kx\} \ne 0$. For $n_0$, there exist $k_0 \in \{1, \dots, n_0\}$ and $\ell_0 \in \mathbb{Z}$ such that
$$0 < k_0 x - \ell_0 < \frac{1}{n_0 + 1}.$$
Without loss of generality, assume that $k_0$ and $\ell_0$ are coprime. Otherwise, we can replace $k_0$ and $\ell_0$ with $\frac{k_0}{\gcd(k_0, \ell_0)}$ and $\frac{\ell_0}{\gcd(k_0, \ell_0)}$ respectively, and the condition still holds.
Let $n_1 = \left\lfloor \frac{1}{k_0 x - \ell_0} \right\rfloor > n_0$. Then there exist $k_1 \in \{1, \dots, n_1\}$ and $\ell_1 \in \mathbb{Z}$ such that
$$0 < k_1 x - \ell_1 < \frac{1}{n_1 + 1} < k_0 x - \ell_0.$$

Similarly, we can assume that $k_1$ and $\ell_1$ are coprime. Since $k_0$ is coprime with $\ell_0$ and $k_1$ is coprime with $\ell_1$, we have $\frac{\ell_0}{k_0} \neq \frac{\ell_1}{k_1}$. Therefore,
$$\begin{aligned} 1 \le |k_0\ell_1 - k_1\ell_0| &= |k_1(k_0x - \ell_0) - k_0(k_1x - \ell_1)| \\ &< \max \{k_1(k_0x - \ell_0), k_0(k_1x - \ell_1)\} \quad (\text{because } k_1(k_0x - \ell_0) > 0, k_0(k_1x - \ell_1) > 0) \\ &\le \max \left\{\frac{k_1}{n_1}, \frac{k_0}{n_1 + 1}\right\} \le 1 \quad (\text{because } \frac{1}{k_0x - \ell_0} \ge n_1, \text{ hence } k_0x - \ell_0 \le \frac{1}{n_1}.) \end{aligned}$$
This leads to a contradiction. $\Box$

*Proof 3.* For any irrational number $x$, the fractional part $\{kx\}$ is distinct and densely distributed in the interval $(0,1)$. Hence, there exist infinitely many positive integers $m$ satisfying:
$$\{mx\} < \{kx\}, \quad \forall k = 1, 2, \dots, m-1.$$
For each such $m \ge 2$, let $\beta$ be the smallest value among $\{x\}$, $\{2x\}$, ..., $\{(m-1)x\}$. Thus, for $k = 1, 2, \dots, m-1$, we have $\{kx\} \ge \beta$, which implies that there are no integers in the open intervals $(kx - \beta, kx)$.
Consider the points $O : (0,0)$, $A : (m,mx)$, $B : (m,mx - \beta)$, $C : (0,-\beta)$ on the coordinate plane. The parallelogram $OABC$ does not contain any lattice points in its interior, and on its boundary, there are exactly three lattice points: $O$, $D : (m, \lfloor mx \rfloor)$, and $E : (r, rx - \beta) = (r, \lfloor rx \rfloor)$.
The lattice triangle $\triangle ODE$ has no lattice points inside or on its boundary, except for the vertices. By Pick's theorem, its area is $\frac{1}{2}$. Therefore, the area of the parallelogram $OABC$ is $\ge 2S_{\triangle ODE} = 1$. On the other hand, the area of this parallelogram is $m \times \beta = m \times \{rx\} \ge 1$. Hence, $\beta \ge \frac{1}{m}$, and for $k = 1, 2, \dots, m-1$, we have
$$\{kx\} \ge \{rx\} \ge \frac{1}{m}.$$
Therefore, $n = m - 1$ satisfies the given condition. There are infinitely many such $n$, implying that no irrational number $x$ satisfies the condition. $\Box$