From now on, I will call highly divisible integers "highlies", for convenience. It is easy to see that the sequence of highlies is infinite and divisible by any number for sufficiently big highlies. To see this, we see $p^k | n$ for a sufficiently big highly $n$, where $p$ is prime. If $k=1$, then $n$ must be a product of a limited number of primes. Say $n=q_1^{e_1}*...*q_l^{e_l}$, where $q_1=2$. We know $e_1$ is bounded, otherwise we can decrease it and multiply $n$ by $q_{l+1}^e$ for an adequate $e$. Therefore $e_i$ is bounded for all $i$, so $n$ is bounded. For $k > 1$, take a big enough $n$, and we will have an arbitrarily large prime divisor of $n$. Divide $n$ by that prime divisor and multiply it by $p^V$ for a big $V$. It is easy to see this new number is smaller than $n$ but has more divisors. So we're done. For part (a), assume there exists infinitely many highlies $H$ such that $kH$ is the next highly, for some $H$. Note that the next highly is simply the least integer $m$ such that $d(m) \ge d(H)$. Clearly $m=2H$ satisfies this. Therefore $k=2$. Now, suppose $H=\prod_{i=1}^k p_i^{e_i}$ is the factorization of $H$. It is easy to see that $p_1=2$, $p_2=3$, etc and also $e_1 \ge e_2 \ge ... \ge e_k$. Now take $m=H\frac{q_{x+1}}{q_x}$ where $q_x$ and $q_{x+1}$ are consecutive prime numbers, and $e_x \textgreater e_{x+1} +1 \ge 1$. By Bertrand's Postulate, $m \textless 2H$. But also $d(m) = d(H) \frac{(e_{x+1}+2)(e_x)}{(e_x+1)(e_{x+1}+1)} \textgreater d(H)$ since $e_x \textgreater e_{x+1}+1$. Therefore, $2H$ isn't the least integer $z$ with $d(z) \textgreater d(H)$, since $m$ also satisfies. So we must have $e_x \le e_{x+1}+1$ for all $x$. If we take $m=4H/3$ it will satisfy, and we only have to care about $6^{V} | H$ where $V$ is a big number I don't want to compute. But that divisibility is true for sufficiently big $H$'s. So we're done. For part (b), choose a $k$ and take a highly $H$ such that $p^k || H$ and WLOG $H$ is the first highly to be divisible by $p^k$. Now suppose $H/p$ isn't highly. Then there exists a highly $x$ such that $d(x) \ge d(H/p)$ and $x \textless H/p$. Now suppose $p^k$ doesn't divide $x$, so say $v_p(x)=q<k$. Then $d(xp) = d(x) \frac{q+2}{q+1} \ge d(H/p)\frac{q+2}{q+1} \ge d(H/p)\frac{k+1}{k} = d(H)$ since $v_p(H/p)=k-1$. So $x$ is a highly divisible by $p^k$ and less than $H$, contradicting the minimality of $H$. So we were wrong and $H/p$ is in fact highly and we're done, since we can choose $k$ from an infinite subset of the naturals.

Here's my solution for part (a). I highly suspect that this might be wrong, so I would be grateful if someone points out any flaws.

Amazing problem!

Yimin Ge wrote: Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$. Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ satisfying $m < s < n$. (a) Show that there are only finitely many pairs of consecutive highly divisible integers of the form $(a, b)$ with $a\mid b$. (b) Show that for every prime number $p$ there exist infinitely many positive highly divisible integers $r$ such that $pr$ is also highly divisible. For brevity, replace the term "highly divisible" with "peak". First note that infinitely many peak numbers exist, otherwise $(d(n))_{n \ge 1}$ is bounded above, which is clearly not the case. Let $q_1<q_2<\dots$ be the sequence of primes. Now we give three important claims. Lemma. All peaks consist of a contiguous block of the first few primes in their factorisation. (Proof) Let $n=\prod^m_{j=1} p_j^{\alpha_j}$ be a peak. Then $d(a)=d(n)$ where $a=\prod^m_{j=1} q_j^{\alpha_j}$ and $a \le n$. So $a=n$ and the claim is true. $\blacksquare$ Lemma. Exponents of these primes in the peak are in decreasing order. (Proof) Suppose $p^a || n$ and $q^b || n$ for $p<q, a<b$ and a peak $n$. Then $d(n)=d\left(\left(\tfrac{p}{q}\right)^{a-b}n\right)$ while the latter is smaller. $\blacksquare$ Lemma. For any prime $q$ and integer $k>0$, all sufficiently large peaks are divisible by $q^k$. (Proof) For $k=1$, suppose for infinitely many peaks $n$ we have $p_{m+1} \nmid n$. Then for a large such $n$ we can find $1 \le s \le m$ with $p_s^{\alpha_s}>\sqrt[m]{n}$. Now we can conveniently drop a few factor of $p_s$ in favour of $p_{m+1}$ to get $n'<n$ but $d(n')>d(n)$, contradiction! Suppose $q=2$ fails for some $k>1$. Then all exponents are small, so we can drop a few factors at the end and append a suitable number of $2$'s. Let $q_{m+1}$ for $m \ge 1$ be the smallest prime that fails for some fixed $k>1$. Then for all peaks $n$ sufficiently large, $v_2(n), v_{p_m}(n)$ exceed a constant $M>0$. Since $q_{m+1}<2q_m$, we see $d\left(\frac{q_{m+1}}{2q_m}n\right)>d(n)$ a contradiction! Hence the lemma is true. $\blacksquare$ Now pick a sufficiently large peak $n$. It is easy to check $\operatorname{max}\left( d(\tfrac{4}{3}n), d(\tfrac{3}{2}n) \right)>d(n)$ so we have a peak $n<k<2n$ solving part I. For part II, fix any large $k$ and let $r$ be the largest peak with $v_p(r)=k$. Let $j<pr$ be the largest peak (so $j>r$). Let $s=v_p(j)>k$. Clearly, $$d(j) \ge d(pr)=\left(\frac{k+2}{k+1}\right)d(r)>\left(\frac{k+2}{k+1}\right)d\left(\frac{j}{p}\right)=\left(\frac{k+2}{k+1}\right)\left(\frac{s}{s+1}\right)d(j)$$hence $(k+2)s<(k+1)(s+1)$ leading to $s<k+1$, that is clearly false! Hence $pr$ is a peak two! $\blacksquare$

Since no one has pointed it out yet, the complete list for part (a) is 1, 2, 6, 12, 60, 360, 2520. http://oeis.org/A072938

Solid problem. Before we go into each part, we first prove a bunch of facts about highly divisible numbers. First, note that there are clearly infinitely many of these numbers, as $d(n)$ can be arbitrarily large, and therefore there are infinitely many peaks. Let $p_1<p_2<\cdots$ be the prime numbers. Lemma: Consider a highly divisible number $n$, and let its prime factorization be $n=\prod_i p_i^{e_i}$, where $e_i=0$ for sufficiently large $i$. We have the following two results. (a) We have $e_1\ge e_2\ge e_3\ge\cdots$. (b) For fixed $t$ and $i$, there exists some $N$ such that $n\ge N\implies e_i\ge t$. Proof: Note that permuting the exponents in the prime factorization does not change the value of $d(n)$, so if the $e_i$s were non increasing, we could permute them to be non increasing, decreasing the value of $n$ while maintaining the value of $d(n)$, thus contradicting highly divisible. This proves (a). Before proving (b), we first show it with $t=1$ (i.e. for fixed $i$, all sufficiently large highly divisible numbers contain $p_i$). Suppose that this was not the case, and that there was some $k$ such that all highly divisible numbers are only divisible by $p_1,\ldots,p_k$. For sufficiently large highly divisible $n=p_1^{e_1}\cdots p_k^{e_k}$, we would have $e_1$ arbitrarily large, and in particular such that \[\log_{p_1}(p_{k+1})<\frac{e_1}{100}.\]Now, pick a positive integer $x$ such that \[\log_{p_1}(p_{k+1})<x<\frac{e_1+1}{2}.\]Now, let $m=\tfrac{n}{p_1^x}p_{k+1}$. We see that $m<n$, and \[\frac{d(m)}{d(n)} = \frac{e_1-x+1}{e_1+1}\ge 1,\]so $n$ is not highly divisible. This proves (b) for $t=1$. Now we'll show (b) for arbitrary $t$. Suppose that there did not exist such $N$, and that we have $e_i<t$ for all highly divisible $n$. Let $u$ be the largest value such that $e_i=u$ for infinitely many highly divisible $n$. Indeed, suppose we have such $n=p_1^{e_1}\cdots p_j^{e_j}$ (with $e_j\ge 1$), so $e_i=u$. Let $m=\tfrac{n}{p_j}\cdot p_i^{2u}$, so $m<n$ (as long as $n$ is large enough, so that $p_j$ is large enough). We see that \[\frac{d(m)}{d(n)} = \frac{e_j}{e_j+1}\cdot\frac{3u+1}{u+1}\ge 1,\]so $n$ is not highly divisible. This is the desired contradiction, so (b) is proved. $\blacksquare$ We'll now solve the original problem. (a) Suppose that $n=p_1^{e_1}\cdots p_k^{e_k}$ is highly divisible, and is sufficiently large. Suppose we have $e_i\ge e_{i+1}+2$ for some $i$. Let $m=\tfrac{n}{p_i}\cdot p_{i+1}$. Then, \[\frac{d(m)}{d(n)} = \frac{e_i}{e_i+1}\cdot\frac{e_{i+1}+2}{e_{i+1}+1}\ge\frac{e_{i+1}+2}{e_{i+1}+3}\cdot\frac{e_{i+1}+2}{e_{i+1}+1}>1,\]and $m<2n$ (by Bertrand). Thus, we have $n<m<2n$ such that $d(m)>d(n)$, so the next highly divisible number will be less than $2n$, so not a multiple of $n$. Now, we have $e_{i+1}\in\{e_i,e_i-1\}$. The idea now is to focus on $i=1$, so $e_2\in\{e_1,e_1-1\}$ (and recall that $e_1$ is sufficiently large by the second part of the lemma). Now, let $m=\tfrac{4n}{3}$, and now \[\frac{d(m)}{d(n)} = \frac{e_1+3}{e_1+1}\cdot\frac{e_2}{e_2+1}\in\{\frac{(e_1+3)(e_1-1)}{e_1(e_1+1)},\frac{e_1(e_1+3)}{(e_1+1)^2}\},\]both of which are greater than $1$, so we finish by the same logic as before. (b) Note that $v_p(n)$ is unbounded as $n$ ranges over highly divisible positive integers (by the second part of the lemma). The key idea is to pick highly divisible $n$ such that $v_p(n)$ is at a peak, or that $v_p(m)<v_p(n)$ for all highly divisible $m<n$. We now show that $r=n/p$ is in fact highly divisible, which finishes since there are infinitely many peaks. Suppose $r$ is not highly divisible. Let $x$ be the largest highly divisible number less than $r$, and since $r$ isn't highly divisible, we must have $d(x)\ge d(r)$. Since $pr$ is highly divisible, we have $d(px)<d(pr)$. Now, we have \[\frac{v_p(x)+2}{v_p(x)+1}d(x) = d(px) < d(pr) = \frac{v_p(r)+2}{v_p(r)+1}d(r),\]so \[\frac{\frac{v_p(r)+2}{v_p(r)+1}}{\frac{v_p(x)+2}{v_p(x)+1}}>1,\]or $v_p(x)>v_p(r)$. Thus, \[v_p(x)\ge v_p(pr)=v_p(n),\]which violates the peak $v_p$ assumption on $n$. This solves (b).

Yimin Ge wrote: Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$. Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ satisfying $m < s < n$. (a) Show that there are only finitely many pairs of consecutive highly divisible integers of the form $(a, b)$ with $a\mid b$. (b) Show that for every prime number $p$ there exist infinitely many positive highly divisible integers $r$ such that $pr$ is also highly divisible. Highly divisible is too long and boring so call them dead integers. Claim: If $n=\prod_i p_i^{e_i}$ is an dead integer then:- (a) We have $e_1\ge e_2\ge e_3\ge\cdots$. (b) $n$ must be divisible by the first $k$ primes. Proof: Note that permuting the exponents in the prime factorization does not change the value of $d(n)$, so if the $e_i$'s were non increasing, we could permute them to be non increasing, decreasing the value of $n$ while maintaining the value of $d(n)$,which is alive a contradiction to the graveyard. For part(b),Let $n=\prod^m_{j=1} p_j^{\alpha_j}$ be dead. Then $d(a)=d(n)$ where $a=\prod^m_{j=1} q_j^{\alpha_j}$ and $a \le n$. So $a=n$ and the claim is true. $\blacksquare$ For part a)Choose a large number which is dead. Then $\operatorname{max}\left( d(\tfrac{4}{3}n), d(\tfrac{3}{2}n) \right)>d(n)$ so there is another dead integer $X$ which is $n<X<2n$,so this is proved. For part b)Choose a large monster $F$ with $\nu_p(n)=F$ Now suppose that $pn$ isn't dead and let $X$ be the smallest dead number which is less than $X<pn$ Now clearly,$$d(j) \ge d(pr)=\left(\frac{k+2}{k+1}\right)d(r)>\left(\frac{k+2}{k+1}\right)d\left(\frac{j}{p}\right)=\left(\frac{k+2}{k+1}\right)\left(\frac{s}{s+1}\right)d(j)$$,so $(k+2)s<(k+1)(s+1)$ or $k>s+1$,another contradiction,so we are done.$\blacksquare$

We first prove a few generic properties of highly divisible numbers. Property 1: There are infinitely many highly divisible numbers. Proof: $d(i)$ is unbounded. $\blacksquare$ Property 2: If $n=p_1^{k_1}\ldots p_n^{k_n}$ is highly divisible, then $k_1 \geq \cdots \geq k_n$. Proof: If we had $k_i<k_{i+1}$ for some $i$, then increase $k_i$ by $1$ and decrease $k_{i+1}$ by $1$, which decreases $n$. Further, this multiplies $d(n)$ by $$\frac{k_i+2}{k_i+1}\cdot \frac{k_{i+1}}{k_{i+1}+1},$$which is at least $1$ as $$\frac{k_i+2}{k_i+1}\cdot \frac{k_{i+1}}{k_{i+1}+1}\geq 1 \iff k_ik_{i+1}+2k_{i+1}\geq k_ik_{i+1}+k_{i+1}+k_i+1 \iff k_{i+1} \geq k_i+1,$$which is true. $\blacksquare$ Property 3: For any prime $p$ and $k\geq 1$, all large enough highly divisible $n$ are divisible by $p^k$. Proof: First, note that if the exponent of some prime $p$ is unbounded, then the exponent of every prime $q \neq p$ is unbounded as well. This is because if the exponent of $q$ is at most $m-1$, we may pick some highly divisible $n$ such that $p^{1001qm} \mid n$. Since $(p^{1000q+q})^{m}>p^{1000qm}q^{m}$, $\frac{nq^m}{p^{qm}}<n$. On the other hand, $\frac{nq^m}{p^{qm}}$ has at least $$\frac{2m}{m}\cdot \frac{1000qm+1}{1000qm+qm+1}>1$$times as many divisors as $n$, hence $n$ is not highly divisible: contradiction. However, if the exponent of every prime $p$ is bounded, then arbitrarily large highly divisible $n$ must be divisible by arbitrarily large primes. But if the exponent of $2$ is bounded by $m$, simply take some prime $p>2^m$ that divides some highly divisible $n$. Then $\frac{n2^m}{p}<n$, yet $\frac{n2^m}{p}$ has at least $$\frac{2m-1}{m-1}\cdot \frac{1}{2}>1$$times as many divisors as $n$: another contradiction. $\blacksquare$ We are now ready to do part (a). The idea here is that if $a,b$ are highly consecutive and $a \mid b$, then $4a/3$ must have more divisors than $a$ if $a$ is large enough. I claim that we have $\nu_2(n)\leq \tfrac{5}{3}\nu_3(n)+\tfrac{16}{3}$ for large highly divisible $n$. Indeed, if this was not the case (so $\nu_2(n)>\tfrac{5}{3}\nu_3(n)+\tfrac{17}{3}$), we could multiply $n$ by $\tfrac{27}{32}$. Letting $\nu_2(n)=x,\nu_3(n)=y$, this multiplies the divisor count of $n$ by $$\frac{x-4}{x+1}\cdot \frac{y+4}{y+1},$$which is at least one, as $$(x-4)(y+4)\geq (x+1)(y+1) \iff xy-4y+4x-16\geq xy+x+y+1 \iff 3x \geq 5y+17$$Then, let $\nu_2(a)=x,\nu_3(b)=y$, so $4a/3$ has $$\frac{x+3}{x+1}\cdot \frac{y}{y+1}$$times as many divisors as $a$. This is greater than $1$ if $$(x+3)y\geq (x+1)(y+1)+1 \iff xy+3y \geq xy+x+y+2 \iff x \leq 2y-2,$$which holds for all $x$ large enough, as $\tfrac{5}{3}y+\tfrac{16}{3}\leq 2y-2$. $\blacksquare$ Now for part (b). By property 3, for any given prime $p$ there exist infinitely many highly divisible $n$ such that $\nu_p(n)<\nu_p(m)$ for all $m>n$ and $m$ highly divisible—call such $n$ extinct. I claim that for all extinct $n$, $pn$ is also highly divisible. Indeed, suppose $pn$ wasn't, so there exists some rational $r<1$ such that $rpn$ is highly divisible and $d(rpn)>d(pn)$. It is clear that we require $rpn>n$. Suppose $\nu_p(r)=k$, so let $r=p^kr'$. If multiplying $n$ by $r'$ multiplies $d(n)$ by some rational $t$, then multiplying $pn$ by $r'$ does the same. Thus, $$\frac{d(rpn)}{d(pn)}=t\cdot \frac{k+\nu_p(n)+2}{\nu_p(n)+2}>1.$$On the other hand, $$\frac{d(rn)}{d(n)}=t\cdot \frac{k+\nu_p(n)+1}{\nu_p(n)+1}.$$If $k \geq 0$, this is at least $t\cdot \frac{k+\nu_p(n)+2}{\nu_p(n)+2}>1$, which contradicts the fact that $n$ is highly divisible. Hence $k<0$, but then $\nu_p(rpn)\leq \nu_p(n)$ and $rpn>n$, hence $n$ cannot be extinct (by taking $m=rpn$ in the definition): contradiction. $\blacksquare$

Wow, surprisingly tough for a 2005 ISL. Let $P = \{2= p_1 < p_2 < p_3 < \ldots\}$ be the set of all primes (a) Let $n$ be some highly divisible number. We can see that $n$ can be represented in the form of $p_1^{e_1}p_2^{e_2} \ldots p_k^{e_k}$ where $e_1 \geq e_2 \geq \ldots \geq e_k$ Claim) For all $l \in \mathbb{N}$, there exists some constant $N_l$ such that all highly divisible numbers $n \geq N_l$ satisfies $v_2(n) > l$

Claim 2) For a sufficiently large highly divisible integer $n$, $v_2(n) - v_3(n) \geq 2$

We can now finish the problem. FTSOC assume there are infinitely many consecutive highly divisible numbers of the form $(n,kn)$ We can see that $d(2n) > d(n)$, and thus $k = 2$. Let $n$ be an arbitrarily large number such that $(n,2n)$ are consecutive highly divisible numbers Note that $\frac{3}{2}n < 2n$, but $(v_2(n))(v_3(n)+2) > (v_2(n)+1)(v_3(n)+1) \implies d(\frac{3}{2}n) > d(n)$ Thus, there must exist a highly divisible number between $n$ and $\frac{3}{2}n$, causing a contradiction (b) We can obviously generalize the first claim to all primes

Let $h_1$, $h_2$, $\ldots$ be the sequence of all highly divisible numbers, and let $p$ be an arbitrary prime By the generalized Claim 1, it is possible to take infinitely many $h_i$ such that $v_p(h_i) < v_p(h_j)$ for all $j > i$ We claim that for all such $i$, $p \cdot h_i$ is also highly divisible