IMO Shortlist

Notice first that $\left\lfloor x+\frac{1}{2}\right\rfloor$ is an integer nearest to $x$, so $\|x\|=\left|\left\lfloor x+\frac{1}{2}\right\rfloor-x\right|$. Thus we have
$$\left\lfloor x+\frac{1}{2}\right\rfloor=x \pm\|x\| .$$
For every rational number $r$ and every prime number $p$, denote by $v_{p}(r)$ the exponent of $p$ in the prime factorisation of $r$. Recall the notation $(2n-1)!!$ for the product of all odd positive integers not exceeding $2n-1$, i.e., $(2n-1)!!=1 \cdot 3 \cdots (2n-1)$.

Lemma. For every positive integer $n$ and every odd prime $p$, we have
$$v_{p}((2n-1)!!)=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor .$$
Proof. For every positive integer $k$, let us count the multiples of $p^{k}$ among the factors $1,3, \ldots, 2n-1$. If $\ell$ is an arbitrary integer, the number $(2\ell-1)p^{k}$ is listed above if and only if
$$0<(2\ell-1)p^{k} \leqslant 2n \quad \Longleftrightarrow \quad \frac{1}{2}<\ell \leqslant \frac{n}{p^{k}}+\frac{1}{2} \quad \Longleftrightarrow \quad 1 \leqslant \ell \leqslant \left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor .$$
Hence, the number of multiples of $p^{k}$ among the factors is precisely $m_{k}=\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor$. Thus we obtain
$$v_{p}((2n-1)!!)=\sum_{i=1}^{n} v_{p}(2i-1)=\sum_{i=1}^{n} \sum_{k=1}^{v_{p}(2i-1)} 1=\sum_{k=1}^{\infty} \sum_{\ell=1}^{m_{k}} 1=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor .$$
$\square$

In order to prove the problem statement, consider the rational number
$$N=\frac{(2a+2b-1)!!}{(2a-1)!!\cdot(2b-1)!!}=\frac{(2a+1)(2a+3) \cdots (2a+2b-1)}{1 \cdot 3 \cdots (2b-1)} .$$
Obviously, $N>1$, so there exists a prime $p$ with $v_{p}(N)>0$. Since $N$ is a fraction of two odd numbers, $p$ is odd.

By our lemma,
$$0<v_{p}(N)=\sum_{k=1}^{\infty}\left(\left\lfloor\frac{a+b}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{a}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{b}{p^{k}}+\frac{1}{2}\right\rfloor\right) .$$
Therefore, there exists some positive integer $k$ such that the integer number
$$d_{k}=\left\lfloor\frac{a+b}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{a}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{b}{p^{k}}+\frac{1}{2}\right\rfloor$$
is positive, so $d_{k} \geqslant 1$. By the earlier formula we have
$$1 \leqslant d_{k}=\frac{a+b}{p^{k}}-\frac{a}{p^{k}}-\frac{b}{p^{k}} \pm \left\|\frac{a+b}{p^{k}}\right\| \pm \left\|\frac{a}{p^{k}}\right\| \pm \left\|\frac{b}{p^{k}}\right\|= \pm \left\|\frac{a+b}{p^{k}}\right\| \pm \left\|\frac{a}{p^{k}}\right\| \pm \left\|\frac{b}{p^{k}}\right\| .$$
Since $\|x\|<\frac{1}{2}$ for every rational $x$ with odd denominator, the relation above can only be satisfied if all three signs on the right-hand side are positive and $d_{k}=1$. Thus we get
$$\left\|\frac{a}{p^{k}}\right\|+\left\|\frac{b}{p^{k}}\right\|+\left\|\frac{a+b}{p^{k}}\right\|=d_{k}=1$$
as required.