Let $x$ be an irrational number (I'd be happy with the answer to the question for a specific choice such as $\pi$). According to
the set $\{\{nx\}\}_{n \in \mathbb{N}}$ is dense in $[0,1]$, where $\{\cdot\}$ is the fractional part. Thus, $$\min\limits_{0<n<N} \{nx\}$$ converges to $0$ for $N\rightarrow\infty$. Is there anything known about the scaling of this series, such as $$\min\limits_{0<n<N} \{nx\} = \mathcal{O}\left(\frac{1}{\operatorname{ln}(N)}\right)$$
Generally we cannot say much more about $$m(N) = m_x(N) := \min_{0 < n < N}\: \lbrace nx\rbrace$$ than $m(N) \to 0$. While for every irrational $x$ there are infinitely many $N$ with $m(N) < \frac{1}{N}$, for every function $f \colon \mathbb{N} \to (0,+\infty)$ with $f(N) \to 0$ we can find (uncountably many) irrational $x$ with $$\limsup_{N \to +\infty} \frac{m_x(N)}{f(N)} = +\infty\,.$$ In that sense, $m_x$ can tend to $0$ arbitrarily slowly. But heuristically the typical behaviour is that $m_x(N)$ doesn't tend to $0$ much more slowly than $\frac{1}{N}$.
To understand $m$ we can use the continued fraction expansion (specifically, the simple continued fraction expansion) of $x$.
Since, as far as I'm aware, we do not know much about the continued fraction expansion of $\pi$ (we do "know" the first several thousand million terms, but not what happens after that), we cannot (yet) rule out that $m_{\pi}(N)$ tends to $0$ very very slowly. But we expect it doesn't.
On the other hand, for every $x$ whose continued fraction expansion has bounded partial quotients (called "coefficients" or "terms" in the wikipedia article), in particular for all quadratic irrationals (these have periodic continued fractions), we have $m_x(N) \asymp \frac{1}{N}$, so things like $m_{\sqrt{2}}$ can be analysed rather well. The continued fraction expansion of $e$ has unbounded partial quotients, but it has a known regular pattern, and we have $m_e(N) \in \mathcal{O}\bigl(\frac{\log N}{N}\bigr)$.
Let's take a look at (simple) continued fractions. The indexing starts with $0$, the $k^{\text{th}}$ convergent to the irrational $x$ with continued fraction expansion $[a_0, a_1, a_2, \dotsc]$ shall be denoted by $p_k/q_k$, the $k^{\text{th}}$ complete quotient $[a_k, a_{k+1}, a_{k+2}, \dotsc]$ by $\alpha_k$.
The first important observation is that the convergents are alternately smaller and larger than $x$, we have $$x - \frac{p_k}{q_k} = (-1)^k\cdot \delta_k$$ with $0 < \delta_k < 1$. (We have much better upper bounds for $\delta_k$, but here I'm only concerned with the sign of the difference.)
Anoher important fact is that the convergents give the best rational approximations to $x$ in a very strong sense:
Let $k > 1$. Then for all positive integers $q < q_{k+1}$ and all integers $p$ we have $$\lvert qx - p\rvert \geqslant \lvert q_k x - p_k\rvert \tag{1}$$ with equality if and only if $p = p_k$ and $q = q_k$.
We define the positive numbers $\varepsilon_k$ by $q_k x - p_k = (-1)^k\varepsilon_k$. From $(1)$ it follows that $$m(q_{2k} + 1) = m(q_{2k+1}) = \varepsilon_{2k}$$ for all $k \geqslant 1$. The recurrence for the convergents together with $\alpha_k = a_k + \frac{1}{\alpha_{k+1}}$ yields \begin{align} \varepsilon_k &= \lvert q_{k}x- p_{k}\rvert \\ &= \Biggl\lvert q_{k}\frac{\alpha_{k}p_{k-1} + p_{k-2}}{\alpha_{k}q_{k-1} + q_{k-2}} - p_{k}\Biggr\rvert \\ &= \frac{\bigl\lvert \alpha_{k}\bigl(p_{k-1}q_{k} - p_{k}q_{k-1}\bigr) + \bigl(p_{k-2}q_{k} - p_{k}q_{k-2}\bigr)\bigr\rvert}{\alpha_{k}q_{k-1} + q_{k-2}} \\ &= \frac{\alpha_{k} - a_{k}}{\alpha_{k}q_{k-1} + q_{k-2}} \\ &= \frac{1}{\alpha_{k+1}\bigl(q_{k} + \frac{q_{k-1}}{\alpha_{k+1}}\bigr)} \\ &= \frac{1}{\alpha_{k+1}q_{k} + q_{k-1}} \\ &= \frac{1}{a_{k+1}q_{k} + q_{k-1} + \frac{q_k}{\alpha_{k+2}}} \\ &= \frac{1}{q_{k+1} + \frac{q_k}{\alpha_{k+2}}} \\ &< \frac{1}{q_{k+1}}\,. \end{align} Thus we have $$m_x(N) < \frac{1}{N}$$ at least for all $N$ such that there is a $k \geqslant 1$ with $q_{2k} < N \leqslant q_{2k+1}$, and of course there are infinitely many such (at least one for each $k$).
On the other hand, between $q_{2k+1}$ and $q_{2k+2}$ bad things can happen. First we note that we always have $$\frac{1}{2q_{k+1}} < \varepsilon_k < \frac{1}{q_{k+1}}$$ and $a_{k+2}q_{k+1} < q_{k+2} = a_{k+2}q_{k+1} + q_k < (a_{k+2} + 1)q_{k+1}$ for $k \geqslant 1$. Also, for $1 \leqslant r \leqslant a_{2k+2}$ we have $$\varepsilon_{2k} > (q_{2k} + rq_{2k+1})x - (p_{2k} + rp_{2k+1}) = \varepsilon_{2k} - r\varepsilon_{2k+1} \geqslant \varepsilon_{2k+2}\,.$$ We see that the denominators $q_{2k} + rq_{2k+1}$ yield new minima for $\{n x\}$ (actually not yet, we also need to consider other $q$ between $q_{2k+1}$ and $q_{2k+2}$, but writing such a $q$ in the form $q_{2k} + rq_{2k+1} + s$ with $0 \leqslant r \leqslant a_{2k+2}$ and $0 \leqslant s < q_{2k+1}$ we can use $(1)$ to see that $\{q x\} > \varepsilon_{2k}$ when $s \neq 0$), but they decrease rather slowly.
Now suppose the partial quotient $a_{2k+2}$ is very large, and pick $r \approx \frac{a_{2k+2}}{2}$. Then for $n = q_{2k} + rq_{2k+1}$ we have $$\{nx\} = \varepsilon_{2k} - r\varepsilon_{2k+1} = \varepsilon_{2k} - \frac{r}{a_{2k+2}}\bigl(\varepsilon_{2k} - \varepsilon_{2k+2}\bigr) \approx \frac{1}{2}\varepsilon_{2k} > \frac{1}{4q_{2k+1}}$$ and $n > rq_{2k+1} > a_{2k+2}$ (since $q_{2k+1} > 2$ for $k \geqslant 1$). Given any $f \in o(1)$ and initial part $[a_0, a_1, \dotsc, a_{2k+1}]$ of a continued fraction, we can always choose $a_{2k+2}$ so large that $$\frac{1}{4 q_{2k+1} f(a_{2k+2})} > e^{k^4}\,,$$ say.
Thus $m_x$ can tend to $0$ slowly if the continued fraction of $x$ has huge even-indexed partial quotients (odd-indexed partial quotients would enter the picture if you considered $\max \:\{nx\}$ or equivalently $\min \:(1 - \{nx\})$ instead of or in addition to $\min \: \{nx\}$).
Usually, however, the partial quotients are small compared to the denominators of the convergents, and if we have $a_{k+1} \leqslant \varphi(q_k)$ for all (sufficiently large) $k$, then we have $$m_x(N) \in \mathcal{O}\biggl(\frac{\varphi(N)}{N}\biggr)\,.$$ For $x$ with bounded partial quotients we can take $\varphi$ as a constant function, and for $e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,\dotsc]$ we have $a_n \ll n$ while $q_n \gg c^n$ for some $c > 1$, whence $a_{k+1} \leqslant K\cdot \log q_k$.
For $\pi = [3,7,15,1,292,1,1,1,2,1,3,1,\dotsc]$ the partial quotients $a_2 = 15$ and $a_4 = 292$ are large relatively to the index, but not so large relatively to the denominators $q_1 = 7$ and $q_3 = 113$. Among the first $20000$ partial quotients there are a few large ones, but relatively to the corresponding denominators $q_k$ they are nevertheless very small. Of course we cannot draw any conclusions from that, but so far, the data we have doesn't indicate that $m_{\pi}$ tends to $0$ slowly.
