A very nice problem, also somewhat obvious motivation

So this isn't as hard as normal N8's?

It's a very good problem, but $||\frac{a}{p^k}||+||\frac{b}{p^k}||+||\frac{a+b}{p^k}||=1$ reminds one of $\lfloor\frac{a}{p^k}\rfloor+\lfloor\frac{b}{p^k}\rfloor+\lfloor\frac{a+b}{p^k}\rfloor$ a lot

Let $x=\left\{ \frac{a}{p^k} \right\} , y=\left\{ \frac{b}{p^k} \right\}$. It suffices to show there exist $p,k$ such that $x,y<\frac 12, x+y>\frac 12$. Let $f(x)=\lfloor 2x \rfloor - 2\lfloor x \rfloor$. This implies that $f\left(\frac{a}{p^k}\right)+f\left(\frac{b}{p^k}\right) < f\left(\frac{a+b}{p^k}\right)$. Consider $$N=\frac{\binom{2a+2b}{a+b}}{\binom{2a}{a}\binom{2b}{b}}$$ If $\nu_p(N)>0$, then $\sum\limits_{k\ge 1} f\left (\frac{a+b}{p^k}\right)-f\left(\frac{a}{p^k}\right)-f\left(\frac{b}{p^k}\right) > 0$, so for some $k$, $f\left(\frac{a+b}{p^k}\right)-f\left(\frac{a}{p^k}\right)-f\left(\frac{b}{p^k}\right)>0$, so we'd be done. By Vandermonde's identity, $\binom{2a+2b}{a+b} = \sum\limits_{c\ge 0} \binom{2a}{c} \cdot \binom{2b}{a+b-c} > \binom{2a}{a} \binom{2b}{b}$. However, $p\ne 2$, so we need $$N=\frac{\binom{2a+2b}{a+b} /2^{\nu_2(\binom{2(a+b)}{a+b}) }}{\binom{2a}{a}\binom{2b}{b} / 2^{\nu_2(\binom{2a}{a}\binom{2b}{b})}}>1$$ it suffices to show $\nu_2(\binom{2(a+b)}{a+b}) \le \nu_2(\binom{2a}{a}\binom{2b}{b})$, which is equivalent to $s_2(a+b) \le s_2(a) + s_2(b) \leftrightarrow \nu_2(\binom{a+b}{a})\ge 0$, which is clear.

WLOG $a < b$. Look at prime power dividing numbers between $2b$ and $2(a+b)$. If any such prime power is at least $2a$, you're done. Now assume otherwise. Then $n$ consecutive odd numbers have all their prime power factors at most $2a$. However, simply look at their product. It follows that by using the elements containing the largest or second or third largest prime and pairing it with another prime power at least 7. (Deal with counter cases with 5, 3 instead.) $\prod_{2 < p \leq 2n} p^{v_p(n!)} > (5n)^{n - \pi(2n) - 2}$ $n! > (10n)^{n - \pi(2n) - 2}$ $n^{\pi(2n) + 2} > (10e)^n$, which is a contradiction for large numbers (starts pretty small) by PNT. (Here we're using that $\pi(n) < \frac{1.3n}{\log(n)}$ for all $n$) Note: This solution, due to its analytic nature, will have ugly details if actually fleshed out. So many mini lemmas for bounding I've omitted other than the tightest part.

Obviously $||x||=\left| x-\left\lfloor x+\frac 12\right\rfloor\right|.$ By easy check for every positive integer $n$ odd prime $q$ holds $v_q((2n-1)!!)=\sum_{i=1}^\infty \left\lfloor \frac{n}{p^i}+\frac 12\right\rfloor,$ and $(2a-1)!!\cdot (2b-1)!!<(2(a+b)-1)!!,$ so there exist prime $p$ satisfying $$v_p((2a-1)!!\cdot v_p((2b-1)!!)<v_p((2(a+b)-1)!!)\implies \sum_{i=1}^\infty \left( \left\lfloor \frac{a+b}{p^i}+\frac 12\right\rfloor -\left\lfloor \frac{a}{p^i}+\frac 12\right\rfloor -\left\lfloor \frac{b}{p^i}+\frac 12\right\rfloor \right)>0.$$Therefore $\exists k\in \mathbb Z ^+:d_k=\left( \left\lfloor \frac{a+b}{p^k}+\frac 12\right\rfloor -\left\lfloor \frac{a}{p^k}+\frac 12\right\rfloor -\left\lfloor \frac{b}{p^k}+\frac 12\right\rfloor \right)>0$. Claim that this choice works. Observe that $1\leq d_k =\pm \left|\left|\frac{a}{p^k}\right|\right|\pm\left|\left|\frac{b}{p^k}\right|\right|\pm\left|\left|\frac{a+b}{p^k}\right|\right|$ $\boxtimes$, but $\left|\left| \frac{1}{2m-1}\right|\right|<\frac 12$ for $m\in \mathbb Z$. Thus $\boxtimes$ can be hold only if all terms in RHS are positive and $d_k=1,$ which completes our proof.