cool problem.

We seek a positive integer N divisible by at least 2024 distinct primes such that the ratios of consecutive divisors are integers. The prime factorization of $N$: $$N = p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_k^{a_k}$$If you select the exponents $a_i$, you can ensure that the ratio of consecutive divisors is always an integer. This can be achieved by arranging the primes in increasing order and the exponents in increasing order. So yeah by choosing a sufficient number of distinct primes and appropriate exponents, we can construct a number N that satisfies the given conditions.

Kalculis wrote: If you select the exponents $a_i$, you can ensure that the ratio of consecutive divisors is always an integer. Um, no? We can never make all individual summands integers unless $N$ is a prime power. (Otherwise, if $p_1<p_2$ are the smallest prime divisors of $N$ and $p_2=d_k$, then $d_{k-1}$ is a power of $p_1$ which of course does not divide $d_k=p_2$.)

straight wrote: cool problem. cool example problemo

Tintarn wrote: Does there exist a positive integer $N$ which is divisible by at least $2024$ distinct primes and whose positive divisors $1 = d_1 < d_2 < \ldots < d_k = N$ are such that the number \[ \frac{d_2}{d_1}+\frac{d_3}{d_2}+\ldots+\frac{d_k}{d_{k-1}} \]is an integer? We construct a sequence $N_1,N_2,\ldots$ of numbers with $1,2,\ldots$ different prime divisors each, recursively in a way that they satisfy the given integrality condition. Start with $N_1=2$. Given $N_n$ choose a prime $p>N_n$ and define $N_{n+1}=N_n\cdot p^{N_n}$. Then the regarded sum $a_{n+1}$ of fractions of divisors is equal to $(N_n+1)a_n+p\in\mathbb{Z}$.

I think that takin a prime $p$ bigger than all the others divisors would be sufficient

The answer is yes Claim : Instead of at least 2024 distinct prime factors, we claim that the problem is correct in general for any natural number n. Proof: we induct on n . For the base case n=1 we only need to put N=2,which is obvious. Now , assume that  for n_1 the answer is K.  Now we ant to create a number that work for n Consider P a prime number bigger than K now for n we write number K*P^k We can easily prove that its works. Our inductive step is complete, so we conclude that this is true for all n