Taiwan IMO 3J, 2020

We construct arbitrarily large pairs $(n, N)$ of integers that meet the conditions of the problem. Fix an integer $x \ge 3$, and let $N = \prod_{p \le x} p$ be the product of the primes at most $x$. We set $n = 2^{2-\pi(x)} \prod_{p \le x} (p-1)$, where $\pi(x)$ denotes the number of primes at most $x$. This $n$ is easily seen to be an even integer, and the pair $(n, N)$ clearly satisfies (b).
To estimate the size of $N$, we compute
$$\frac{4N}{2^{\pi(x)}n} = \prod_{p \le x} \frac{p}{p-1} = \prod_{p \le x} (1 + p^{-1} + p^{-2} + \dots) > \sum_{r=1}^{x!} \frac{1}{r} = \log(x!) + O(1),$$
using the well-known approximation $\sum_{r=1}^{m} \frac{1}{r} = \log(m) + O(1)$ for harmonic numbers.
By definition $x! \ge N \ge n$, so rearranging we obtain the inequality
$$N > \frac{1}{4} \cdot 2^{\pi(x)} n \log(n) + O(2^{\pi(x)} n)$$
and so we are done if we choose $c < 1/4$.

As in the preceding solution, we seek arbitrarily large pairs of integers $(n, N)$ that meet the conditions of the problem. This time, to construct such pairs, we fix a positive integer $t$ and choose distinct prime numbers $p|2^{2^{t-1}} + 1$ and $q|2^t + 1$; we set $N = pq$. It is well-known that $2^t|p-1$ and $2^{t+1}|q-1$, hence
$$n = \frac{(p-1)(q-1)}{2^t}$$
is an integer and the pair $(n, N)$ satisfies (b).