Problem 1847

$n$ is divisible by $\varphi\left(p^{e}\right)$ for every prime power $p^{e}$ exactly dividing $N$.
This property ensures that all $n^{\text{th}}$ powers are congruent to $0$ or $1$ modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\text{th}}$ powers is congruent to $m$ or $m-1$ modulo $p^{e}$, since at most one of the $n^{\text{th}}$ powers is divisible by $p$. Thus, if $k$ denotes the number of distinct prime factors of $N$, we find by the Chinese Remainder Theorem at most $2^{k} m$ residue classes modulo $N$ which are sums of at most $m$ pairwise coprime $n^{\text{th}}$ powers. In particular, if $N>2^{k} m$ then there are infinitely many positive integers not expressible as a sum of at most $m$ pairwise coprime $n^{\text{th}}$ powers.
It thus suffices to prove that there are arbitrarily large pairs $(n, N)$ of integers satisfying $(\dagger)$ such that
$$N>c \cdot 2^{k} n \log (n)$$
for some positive constant $c$.
We construct such pairs as follows. Fix a positive integer $t$ and choose (distinct) prime numbers $p \mid 2^{2^{t-1}}+1$ and $q \mid 2^{2^{t}}+1$; we set $N=p q$. It is well-known that $2^{t} \mid p-1$ and $2^{t+1} \mid q-1$, hence
$$n=\frac{(p-1)(q-1)}{2^{t}}$$
is an integer and the pair ( $n, N$ ) satisfies ( $\dagger$ ).
Estimating the size of $N$ and $n$ is now straightforward. We have
$$\log _{2}(n) \leqslant 2^{t-1}+2^{t}-t<2^{t+1}<2 \cdot \frac{N}{n}$$
which rearranges to
$$N>\frac{1}{8} \cdot 2^{2} n \log _{2}(n)$$

Solution 2, obtaining better bounds. As in the preceding solution, we seek arbitrarily large pairs of integers $n$ and $N$ satisfying $(\dagger)$ such that $N>c 2^{k} n \log (n)$.
This time, to construct such pairs, we fix an integer $x \geqslant 4$, set $N$ to be the lowest common multiple of $1,2, \ldots, 2 x$, and set $n$ to be twice the lowest common multiple of $1,2, \ldots, x$. The pair $(n, N)$ does indeed satisfy the condition, since if $p^{e}$ is a prime power divisor of $N$ then $\frac{\varphi\left(p^{e}\right)}{2} \leqslant x$ is a factor of $\frac{n}{2}=\operatorname{lcm}_{r \leqslant x}(r)$.
Now $2 N / n$ is the product of all primes having a power lying in the interval ( $x, 2 x$ ], and hence $2 N / n>x^{\pi(2 x)-\pi(x)}$. Thus for sufficiently large $x$ we have
$$\log \left(\frac{2 N}{2^{\pi(2 x)} n}\right)>(\pi(2 x)-\pi(x)) \log (x)-\log (2) \pi(2 x) \sim x$$
using the Prime Number Theorem $\pi(t) \sim t / \log (t)$.
On the other hand, $n$ is a product of at most $\pi(x)$ prime powers less than or equal to $x$, and so we have the upper bound
$$\log (n) \leqslant \pi(x) \log (x) \sim x$$
again by the Prime Number Theorem. Combined with the above inequality, we find that for any $\epsilon>0$, the inequality
$$\log \left(\frac{N}{2^{\pi(2 x)} n}\right)>(1-\epsilon) \log (n)$$
holds for sufficiently large $x$. Rearranging this shows that
$$N>2^{\pi(2 x)} n^{2-\epsilon}>2^{\pi(2 x)} n \log (n)$$
for all sufficiently large $x$ and we are done.