Assume not. Then there exists $N$ such that for all $n>N$, we have $\{nx\} > \frac{1}{n+1}$, since that implies any $n' > \frac{1}{\min_{1\le j\le N} \{jx\}}$ satisfy $\min_{1\le k\le n'} \{ kx \} > \frac{1}{n'+1}$. By Dirichlet's theorem, $$\min_{1\le k\le n} \{ ||kx|| \} < \frac{1}{n+1}$$ Let $S = \{ k \mid ||kx|| < \frac{1}{k+1} \}$. By Dirichlet's theorem, $S$ is infinite, and by our assumption for all $k\in S$, we have $\{kx\} > 1-\frac{1}{k+1}$. Let $b_k = 1-\{kx\}$. Let $k$ be a large positive integer satisfying $$||kx|| = \min_{1\le m\le k} ||mx||$$and $\frac{1}{k+1} < \frac{1}{\min_{1\le j\le N} \{jx\}}$ I will induct on $t$ to show that for any $t\in \mathbb{N}$, $$||kx|| < \frac{1}{tk+1}$$ Which clearly finishes the problem. Base Case: $t=2$. I will show that $$||kx|| = \min_{1\le m\le 2k} ||mx||$$ Assume for contradiction $b_{p} < b_{k}$ for some $p\in [k+1,2k]$. Then $$\{(p-k)x\} = \{px\} - \{kx\} = b_{k}-b_{p} \in \left[0, \frac{1}{k+1}\right]$$Thus $\{(p-k)x\}\le \frac{1}{k+1} \le \frac{1}{p-k+1} $ This implies that $p-k\in S$. When $\{(p-k)x\} <\frac{1}{k+1}$, then $p-k>N$ or we have contradicted the assumption that $\frac{1}{k+1} < \frac{1}{\min_{1\le j\le N} \{jx\}}$. However, now we have also contradicted our assumption that for all $n>N, \{nx\} > \frac{1}{n+1}$. Inductive step: I will show that $$||kx|| = \min_{1\le m\le tk} ||mx|| \le \frac{1}{tk+1}$$ By our inductive hypothesis, $$||kx|| < \frac{1}{(t-1)k+1}$$ Assume there exists $p\in [(t-1)k+1, tk]$ such that $||px|| < ||kx|| \iff b_k>b_p$. Then $$\{(p-(t-1)k)x\} \equiv (t-1)b_k-b_p (\bmod\; 1)$$ Since $b_k\ge b_p$, we can see that $0\le (t-1)b_k-b_p \le (t-1)b_k < \frac{t-1}{(t-1)k+1}$. If $p-(t-1)k < k$ we are done because $\frac{t-1}{(t-1)k+1} < \frac 1k \le \frac{1}{p-(t-1)k+1}$. If $p=tk$, then we have $1-\frac{1}{k+1} < \{kx\} < \frac{t-1}{(t-1)k+1}$, which is absurd.

Note: The problem asks us to prove that infinitely many $n$ satisfy $\min_{1\le k\le n} \{kx\} > \frac{1}{n+1}$. The above post proves that infinitely many $n$ satisfy $\min_{1\le k\le n} \{kx\} < \frac{1}{n+1}$

Assume for contradiction for all $n>N$, we have $\min_{1\le k\le n} \{kx\} < \frac{1}{n+1}$. Suppose $\{nx\} = \min_{1\le k\le n} \{kx\}$. Let $\{lx\} = \min_{1\le k\le n-1} \{kx\}$. Then note that $\{nx\} < \{lx\} < \frac 1n$. Finally, $$\{nlx\} = l\{nx\} = n\{lx\}$$ Since $\{nx\}, \{lx\} < \frac 1n$. Since $l<n, \{nx\} < \{lx\}$, this is absurd!

The answer is no. Indeed, for any positive irrational number $x$, there are infinitely many $n\in\mathbb{Z}^+$ such that $\{kx\}\ge\frac{1}{n+1}$ for $1\le k\le n,$ where $k\in\mathbb{Z}^+$. For a positive irrational number $x$, consider the set $$S_x\coloneq\{n: \{kx\}\geq \frac{1}{n+1},\,\, 1\le k\le n;\,\, k, n\in\mathbb{Z}^+\}.$$For the sake of contradiction, assume there exists a positive irrational number $x_0$ such that $S_{x_0}$ is finite. Note that $$\{kx_0\}\ge \frac{1}{1+\max{S_{x_0}}},\,\, 1\le k\le \max{S_{x_0}}.$$Due to our assumption, for $n=i+\max{S_{x_0}}$, where $i=1,2,\dots,$ there must exist a non-empty subset of $[1+\max{S_{x_0}}, i+\max{S_{x_0}}]$, say $K_i$, such that for $k\in K_i$, we have $\{kx_0\}<\frac{1}{1+i+\max{S_{x_0}}}$. Let $k_i=\min{K_i}$. It is easy to see that $k_1=1+\max{S_{x_0}}$. As $i$ gets large, $\frac{1}{1+i+\max{S_{x_0}}}$ will eventually be smaller than $\{(1+\max{S_{x_0}})x_0\}$. Thus, there are positive integers $h, j\ge 2$ such that $$k_1=\cdots=k_{j-1}=1+\max{S_{x_0}}<k_j=h+\max{S_{x_0}}\le j+\max{S_{x_0}}.$$From this, we infer that $$\{k_{j-1}x_0\}<\frac{1}{j+\max{S_{x_0}}} \implies \{(1+\max{S_{x_0}})x_0\}<\frac{1}{j+\max{S_{x_0}}}\le\frac{1}{h+\max{S_{x_0}}}\implies (h+\max{S_{x_0}})\{(1+\max{S_{x_0}})x_0\}<1$$and $$\{k_jx_0\}<\frac{1}{1+j+\max{S_{x_0}}} \implies \{(h+\max{S_{x_0}})x_0\}<\frac{1}{1+j+\max{S_{x_0}}}<\frac{1}{1+\max{S_{x_0}}}\implies (1+\max{S_{x_0}})\{(h+\max{S_{x_0}})x_0\}<1.$$ Lemma. If $0<x\{yz\},\,\,y\{xz\}<1$, where $x, y\in\mathbb{Z}^+$, then $\{xyz\}=x\{yz\}=y\{xz\}$. Proof. Since $x\{yz\}=[x\{yz\}]+\{x\{yz\}\}=\{x\{yz\}\},$ we have $$\{xyz\}=\{x([yz]+\{yz\})\}=\{x\{yz\}\}=x\{yz\}.$$Again, since $y\{xz\}=[y\{xz\}]+\{y\{xz\}\}=\{y\{xz\}\},$ we have $$\{xyz\}=\{y([xz]+\{xz\})\}=\{y\{xz\}\}=y\{xz\}.\,\, \blacksquare$$ But this suggests that $$(1+\max{S_{x_0}})\{(h+\max{S_{x_0}})x_0\}=(h+\max{S_{x_0}})\{(1+\max{S_{x_0}})x_0\} \implies \{(h+\max{S_{x_0}})x_0\}=\frac{(h+\max{S_{x_0}})\{(1+\max{S_{x_0}})x_0\}}{1+\max{S_{x_0}}}>\{(1+\max{S_{x_0}})x_0\},$$which is a contradiction with the fact that $$\{(1+\max{S_{x_0}})x_0\}\ge\frac{1}{1+j+\max{S_{x_0}}}>\{(h+\max{S_{x_0}})x_0\}.$$Hence, our assumption is false, and $S_x$ is always infinite for any positive irrational number $x.\,\, \blacksquare$