Such a sequence exists, and a sketch is as follows. Notice first that, for any $m\neq 1$, $d(a_m,a_1)=d(m,1)=0$, and thus, all the terms are coprime to $a_1$. With this, let $q_1\geqslant 2018^{2018}$ be a prime, and assign $a_1=q_1$. Now, if $m=p,n=q$ primes, then. observe that, $d(a_p,a_q)=d(p,q)=0$. Thus, it is tempting to make $a_p$ to be a power of a prime, for each prime $p$. Take a sequence $2018^{2018}<q<q_1<q_2<\cdots$ of primes, all greater than $2018^{2018}$, and let $a_{p_k}=q_k^{n_k}$, where the exponent is $n_k$ will be used to control the monotonicity condition. Now, for each $n\geqslant 2$, we set $a_n$ to be: $$ a_n = \prod_{p_k\text{ prime }:p_k\mid n}q_k^{e_{k,n}}, $$where the exponents are to be determined. Observe that, this sequence indeed obeys the first and the third conditions. Furthermore, as long as $n_k,e_{k,n}\geqslant 1$ always, we indeed retain $d(m,n)=d(a_m,a_n)$ condition. For adjusting the exponents, first fix an arbitrary exponent for $a_2$ and $a_3$. Now for $a_4$, if $a_2=q_2^{n_2}$, then make $a_4=q_2^{e_{2,4}}$ such that, $a_4>a_2,a_3$ is satisfied. Then adjust $a_5$, then $a_6$, and continue in this manner. Keeping in mind that we are free to adjust the exponents, this procedure can be continued as long as we want to.

Let $\{q\}$ be the sequence consisting of all primes greater than $2018^{2018}$ and $\{p\}$ be the sequence of all primes as usual. Set $a_1=2018^{2018}$, and for each $n=\prod\limits p^{e_{k,n}}_{i_{k,n}}$, construct $a_n$ as follow: $a_n=(\prod\limits_{k=1}^{\omega(n)-1} q^{e_{k,n}}_{i_{k,n}})(q_{\omega(n),n}^{E_n})$ with $E_n$ selected so that $a_m\le a_n$ for all $m\le n$. It's easy to see that this construction does satisfy the constraints given by the problem.

Such a sequence does exist. First, we provide the construction. Let $a_1$ be the smallest prime larger than $2018^{2018}$. For $n\geq 2$, define $a_n$ recursively as follows: if $n$ is prime, let $a_n$ be the smallest prime larger than $a_{n-1}$, and if $n = pq$ for coprime $p$, $q > 1$ let $a_n = (a_pa_q)^N$ , where $N$ is sufficiently large such that $a_n > a_{n-1}$. This clearly satisfies the first and second properties. For the third, notice that a prime $s$ divides $a_n$ iff $s = a_t$ and $t | n$ for a prime $t$, so the number of distinct prime divisors of $a_n$ equals that of $n$, so $d(m, n) = d(a_m, a_n)$ always holds.

XbenX wrote: For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60, 126) = d(2^2 \cdot 3 \cdot 5, 2 \cdot 3^2 \cdot 7) = 2.$ Does there exist a sequence $(a_n)$ of positive integers such that: $a_1 \geq 2018^{2018};$ $a_m \leq a_n$ whenever $m \leq n$; $d(m, n) = d(a_m, a_n)$ for all positive integers $m\neq n$? (Dominic Yeo, United Kingdom) Notice that $d(m,n)=\omega(\gcd(m,n))=\omega(\gcd(a_m,a_n))$ Define $$\mathcal{Q}:=\{q_1,q_2,................,q_k\} \;\;\;\;\;\;\;\;\;\;\;\; q_1>2018^{2018}$$Now choose $a_1=q_1$ and for all $m$ \begin{align*} a_m \equiv 0 \pmod {a_{d_1}^{r_1}} \\.\\.\\.\\.\\.\\.\\.\\.\\ a_m \equiv 0 \pmod {a_{d_{\tau(m)}}^{r_{\tau(m)}}} \;\;\;\; d_i|m \end{align*}Clearly $a_i$ satisfy the condition.