Two problems about Squarefree numbers

By the inclusion-exclusion principle, the number of squarefree numbers not exceeding $N$ is

$$\begin{align} \operatorname{sf}(N) &= \sum_{k = 0}^\infty \mu(k)\left\lfloor\frac{N}{k^2}\right\rfloor\\ &= N\sum_{k=0}^\infty \frac{\mu(k)}{k^2} - \sum_{k \leqslant\sqrt{N}}\mu(k)\left(\frac{N}{k^2} - \left\lfloor\frac{N}{k^2}\right\rfloor\right) - \sum_{k > \sqrt{N}} \mu(k)\frac{N}{k^2}\\ &= \frac{6}{\pi^2}N - \sum_{k \leqslant\sqrt{N}}\mu(k)\left(\frac{N}{k^2} - \left\lfloor\frac{N}{k^2}\right\rfloor\right) - \sum_{k > \sqrt{N}} \mu(k)\frac{N}{k^2}\\ &\geqslant \frac{6N}{\pi^2} - 2\sqrt{N}, \end{align}$$

where $\mu$ is the Möbius function.

For $N > 1192$, that bound implies $\operatorname{sf}(N) \geqslant \frac{11}{20}N$. For $N \leqslant 1192$, this bound can easily be verified by explicit computation.

Since the smallest (non-negative) squarefree number is $1$, to write $0$ or $1$ as the sum of two squarefree numbers, we need to allow negative squarefree numbers. If we do, we have $0 = 1 + (-1)$ and $1 = 2 + (-1)$.

For $n \geqslant 2$, the above bound implies that $n - S(n)$ and $S(n)$ intersect, where $S(n) = \{ k : 0 \leqslant k \leqslant n, k \text{ squarefree}\}$, so $n$ can be written as the sum of two non-negative squarefree numbers.

For the difference, $0 = 1 - 1$, and for positive $n$, we set $N = 20n$, and find that there are at least $10n$ squarefree numbers $n < k \leqslant N$, and at least $\left\lceil \frac{209}{20}n\right\rceil > 10n$ squarefree numbers $\leqslant 19n$, hence the sets $S(20n)$ and $n + S(19n)$ must intersect, since there are only $19n$ numbers between $n$ and $20n$, hence $n$ can be written as the difference of two squarefree numbers.

Hint for the sum version: if $n$ cannot be represented as a sum of squarefree numbers, then at least half of the numbers less than $n$ must be squareful. But the limit density of squarefree numbers is $\frac{6}{\pi^2} \gt 0.5$, so already we have that there can only be a finite number of exceptions.