Nice Problem! I take $d$ to be the number amongst those given which is congruent to $15 \bmod 20$. In particular it is a multiple of $5$, and since it is congruent to $3 \bmod 4$ it has a divisor $p$ of the form $3 \bmod 4$. Surely $d$ cannot be a perfect square (as it is congruent to $3 \bmod 4$). So given any natural number $n$, the number $n^2d$ is not a perfect square and so I can find a positive integer $m$ such that $m^2 < n^2d < (m+1)^2$. Then \[\{n\sqrt{d}\} = n\sqrt{d} - m = \frac{n^2d-m^2}{n\sqrt{d} + m} > \frac{n^2d-m^2}{2n\sqrt{d}}\] and so \[ n\sqrt{d}\{n\sqrt{d}\} > \frac{nd^2-m^2}{2} \geqslant \frac{5}{2}\] The reason the very last inequality holds is the following: $n^2d \neq m^2+1$ because the left hand side is a multiple of $p$ while the right hand side is not. (As $-1$ is not a perfect square $\bmod p$). $n^2d \neq m^2+2$ because the left hand side is a multiple of $5$ while the right hand side is not. $n^2d \neq m^2+3$ because the left hand side is a multiple of $5$ while the right hand side is not. $n^2d \neq m^2+4$ because the left hand side is a multiple of $p$ while the right hand side is not. (As $-1$ is not a perfect square $\bmod p$ and thus $-4$ is also not a perfect square $\bmod p$).

This is a way to rationalize looking at numbers that are $15$ modulo $20.$ The condition $n\sqrt{d}\lbrace n \sqrt{d} \rbrace>5/2$ is not very pleasing, so we try to see what it means for $n=1$ (which might also give us a general view, since we may replace $(d,n)$ with $(n^2d,1)$). We prove a lemma: Lemma: For any $x\in \mathbb{N},$ $\sqrt{x}\lbrace \sqrt{x} \rbrace > 5/2$ if and only if $x=a^2+b$ with $(a,b)\in \mathbb{N}^2$ such that $5\leq b \leq 2a.$ Proof. Fix $x\in \mathbb{N}.$ First, note that $\sqrt{x} \lbrace \sqrt{x} \rbrace>5/2$ implies that $x>2$ and $x$ is nonsquare. So, assume $x>2$ is a nonsquare. Write $x=a^2+b$ with $(a,b)\in \mathbb{N}^2$ and $ b \leq 2a.$ Then, $\sqrt{x}\lbrace \sqrt{x} \rbrace>5/2$ if and only if $a^2+b- a\sqrt{a^2+b} > 5/2$ if and only if $a^2+b-5/2 > \sqrt{(a^2)^2 + a^2b}$ if and only if (squaring both sides) $(b-5/2)^2>(5-b)a^2.$ If $ b\leq 4$ then $(b-5/2)^2/(5-b)\leq (b-5/2)^2\leq 9/4 < 4 \leq (2b)^2,$ so we can't have $\sqrt{x}\lbrace \sqrt{x} \rbrace>5/2.$ On the other hand, if $b\geq 5$ then any such $x$ satisfies $\sqrt{x}\lbrace \sqrt{x} \rbrace>5/2.$ $\Box$ Let's apply the lemma to $x=n^2d$ (for $d$ nonsquare). For any $(n,d)\in \mathbb{N}^2,$ we may write $n^2d=a_{n,d}^2+b_{n,d}$ for $(a_{n,d},b_{n,d})\in \mathbb{N}^2$ with $b_{n,d} \leq 2a_{n,d}.$ According to the lemma, the problem asks for a sequence $d_1<d_2<\cdots$ of naturals such that, for each $k,$ $d_{k+1}-d_k\leq 20$ and also $\min_{n,k} \lbrace b_{n,{d_k}} \rbrace\geq 5.$ Taking $n^2d=a_{n,d}^2+b_{n,d}$ modulo $d,$ we see that the condition $b_{n,d}\geq 5$ is satisfied if $d$ is nonsquare and $-1,-2,-3,-4$ are quadratic nonresidues modulo $d.$ In particular, this is true for any $d$ satisfying $d\equiv 3 \pmod{4}$ and $d\equiv 0 \pmod{5},$ i.e., for $d\equiv 15 \pmod{20}.$ $\Box$ Alternatively, after the lemma, one sees that if the $20$ numbers we start with are $1,\cdots,20,$ then, according to the lemma, the number we should choose is either $14$ or $15.$ Then, one can show that the set $\left( \lbrace 15k \rbrace_{k\in \mathbb{N}} \setminus \lbrace 15^2\ell^2 \rbrace_{\ell\in \mathbb{N}}\right) \cup \lbrace 15^2\ell^2\pm 10 \rbrace_{\ell\in 2\mathbb{N}-1} \cup \lbrace 15^2\ell^2-5 \rbrace_{\ell\in 2\mathbb{N}}$ gives a solution. In detail: Suppose $d=15k$ for some $k\in \mathbb{N}$ with $k\neq 15\ell^2$ for any $\ell \in \mathbb{N}.$ Then, for any $n\in \mathbb{N},$ there are $(a_n,b_n)\in \mathbb{N}^2$ such that $n^2d=a_n^2+b_n$ and $1\leq b_n \leq 2a_n.$ Further, we can't have $1\leq b_n \leq 4,$ for the only quadratic residues modulo $15$ are $\lbrace 0,1,4,6,9,10 \rbrace.$ Hence, $n\sqrt{d} \lbrace n \sqrt{d} \rbrace>5/2$ for any $n\in \mathbb{N}.$ Now, suppose $d=15k+\delta$ for $k=15\ell^2$ with $\ell \in \mathbb{N}$ and $\delta \in \lbrace -14,\cdots,-1,1,\cdots,14 \rbrace.$ We want to show that, for any $\ell,$ there is always a choice of $\delta$ such that $n\sqrt{d}\lbrace n \sqrt{d} \rbrace>5/2$ for any $n\in \mathbb{N}.$ If $\ell$ is odd, we may choose $\delta \in \lbrace \pm 10 \rbrace,$ and if $\ell$ is even we may set $\delta=-5.$ Then, $5\mid d$ and $d\equiv 3 \pmod{4}.$ Then, $d$ is nonsquare and $-1,-2,-3,-4$ are quadratic non-residues modulo $d.$ Hence, if $n^2d =a_n^2+b_n$ with $0\leq b_n \leq 2a_n,$ then $5\leq b_n \leq 2a_n.$ By the lemma, we get the desired result.

Very artificial, notice that $n\sqrt{d} \{ n \sqrt{d} \} = \frac{n\sqrt{d}}{n\sqrt{d} + \lfloor n\sqrt{d} \rfloor}*(n^2d - (\lfloor n \sqrt{d} \rfloor)^2) > 5/2$ if we can prove that $n^2d-m^2$ is not in $\{ 0,1,2,3,4 \}$ for all $n,m$. After using mod 5 and 4 (which are suggested by the 20 in the problem), we get that 15 mod 20 works. I also got that if $15 | d$ and $d$ isn't a square, it also works.

Demetres wrote: The reason the very last inequality holds is the following: $n^2d \neq m^2+1$ because the left hand side is a multiple of $p$ while the right hand side is not. (As $-1$ is not a perfect square $\bmod p$). ... $n^2d \neq m^2+4$ because the left hand side is a multiple of $p$ while the right hand side is not. (As $-1$ is not a perfect square $\bmod p$ and thus $-4$ is also not a perfect square $\bmod p$). Forgive me if I am ignorant, but can someone explain these steps? I am not very good... Thank you very much

It is a well-known fact that if $p \equiv 3 (mod 4)$ is a prime and $p | a^2 + b^2$, then $p|a$ or $p|b$.

zschess wrote: It is a well-known fact that if $p \equiv 3 (mod 4)$ is a prime and $p | a^2 + b^2$, then $p|a$ or $p|b$. but how do you know that $p|m$? (or I'm analyzing it wrong )

Sorry but the way I solved it, there are more than one $d$ among $20$ consecutive positive integers... First, prove bu induction that among any $20$ consecutive positive integers there is a number $d$ in the form $d=a^2+b$ where $6\leq b < 2a+1$ Note that, $(na)^2 < n^2(a^2+b) > (na+1)^2 $ therefore $ na < \sqrt{n^2(a^2+b)} < na+1 $ which implies that $ \lfloor \sqrt{n^2(a^2+b)} \rfloor = na$ Also note that $n\sqrt{d} \{ n \sqrt{d} \} = n^2 d - n \sqrt{d}* \lfloor \sqrt{n^2(d)} \rfloor =n^2 d- n \sqrt{a^2+b}*na= n^2(a^2+b - \sqrt{a^2+b}*a) >5/2$ The last inequality is because $n \geq 1$ and after some easy calculations, using the fact that $ b>5$ we get the desired result

My solution is basically the same as others, but I will add a more elaborated solution. Claim. For $d\equiv 15 \ (mod \ 20)$, the inequality always hold for all $n\in \mathbb{N}$. Proof. First, we have $$n\sqrt{d} \left\{ n\sqrt{d} \right\} = n\sqrt{d} \left( n\sqrt{d}- \lfloor n\sqrt{d} \rfloor \right)$$Now, because $n\sqrt{d}$ is not an integer. Therefore, $$n\sqrt{d} \left\{ n\sqrt{d} \right\} = n\sqrt{d} \left( n\sqrt{d} - \lfloor n\sqrt{d} \rfloor \right) > \left( \frac{n\sqrt{d} +\lfloor n\sqrt{d} \rfloor }{2} \right) \left( n\sqrt{d} - \lfloor n\sqrt{d} \rfloor \right) = \frac{n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2}{2}$$Clearly $n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2$ is a positive integer, so it suffices to prove $n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2 \neq 1,2,3,4$. If $n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2=1$, then $n^2d\equiv 1,2 \ (mod \ 4) \Rightarrow n^2\equiv 2,3 \ (mod \ 4) $. A contradiction. If $n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2=2$, then $\left( \lfloor n\sqrt{d} \rfloor \right)^2\equiv 3\ (mod \ 5)$. A contradiction. If $n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2=3$, then $\left( \lfloor n\sqrt{d} \rfloor \right)^2\equiv 2\ (mod \ 5)$. A contradiction. If $n^2d- \left( \lfloor n\sqrt{d} \rfloor \right)^2=4$, then $n^2d\equiv 0,1 \ (mod \ 4) \Rightarrow n^2\equiv 0,3 \ (mod \ 4) $. Thus, $n$ and $\lfloor n\sqrt{d} \rfloor$ are even. Let $n=2a$ and $\lfloor n\sqrt{d} \rfloor =2b$ with positive integers $a,b$. Now we have, $$4a^2d-4b^2=4 \Rightarrow a^2d-b^2=1 \Rightarrow a^2d\equiv 1,2 \ (mod \ 4) \Rightarrow a^2\equiv 2,3 \ (mod \ 4)$$A contradiction, having all cases exhausted leads to the desired conclusion.

Nice one. We claim that $d \equiv 15$ (mod $20$) holds. First note that $d \equiv 3$ (mod $4$), so $d$ is not a square. Next we show that $-1,-2,-3,-4$ are QNRs mod $d$. Note that the exists a prime $p$ dividing $d$ such that $p \equiv 3$ (mod $4$). Since $-1$ and $-4=(-1)2^2$ are QNRs mod $p$, they are also QNRs mod $d$. Also, $5 \mid d$, so $-2, -3$ cannot be QRs mod $d$. Now, note that $\lfloor n\sqrt{d} \rfloor = \lfloor \sqrt{dn^2} \rfloor \leq \sqrt{dn^2-5}$ since $dn^2-a$ cannot be a square for $a=0,1,2,3,4$. So $\{ n\sqrt{d} \} \geq \sqrt{dn^2}-\sqrt{dn^2-5} = \frac{5}{\sqrt{dn^2}+\sqrt{dn^2-5}} > \frac{5}{2\sqrt{dn^2}}$. It follows that $n\sqrt{d} \{ n\sqrt{d} \} > \frac{5}{2}$, so we are done.