That's not really hard but I still get stucked for hours. WLOG, suppose $p < q < 2p$. $\boxed{\textbf{Lemma.}\; \text{Given natural numbers}\; m > 1, n\; \text{where}\; \gcd(m, n) = 1,\; \text{there exists unique positive integer} \; 1\le n < m \;\text{satisfy}\; a.n\equiv 1\pmod{m}}$

Sketch

Since $p$ and $q$ are prime numbers, we have $\gcd(p, q) = 1$ then the lemma tells us that there is $1\le a < q$ satisfy $a.p \equiv 1\pmod{q}$ Case 1. $1\le a \le p$ then consider two consecutive natural numbers: $a.p$ and $a.p - 1$. Here, $1\le a\le p$ hence $a.p$ has the largest prime divisor which is $p$. And $a.p - 1 < q^{2} \implies \frac{ap - 1}{q} < q \in \mathbb{Z}^{+}$ means all prime divisors of $ap - 1$ doesn't exceed $q$. We're done. Case 2. $p < a < q$. Note that $(q - a).p \equiv -1\pmod{q}$ and $1\le q - a < q$ Consider $(q - a)p$ and $(q - a)p + 1$. Firstly, we have $p\mid (q - a)p$ and $q\mid (q - a)p + 1$. On other hand, $q - p < p < a \implies q - a < p$ then the largest prime divisor of $(q - a)p$ is $p$. And $(q - a)p + 1 < q^{2} \implies \frac{(q - a)p + 1}{q} < q$ and we're done. What a beautiful problem

Great solution!

WLOG, assume $1 < \frac{p}{q} < 2$. If $q=2$ then $p=3$, for this case just choose $8$ and $9$. If $q=3$, then $p=5$ and just choose $5$ and $6$. Now assume $q \geq 5$. Let $1 \leq t \leq q-1$ be such that $t \equiv \frac{-1}{p} \pmod{q}$. Now we consider two cases. If $t$ is odd, then $\frac{pt + 1}{2q}$ is an integer. Furthermore we have $\frac{pt + 1}{2q} < \frac{2qt + 1}{2q} = t + \frac{1}{2q}$, implying that $\frac{pt + 1}{2q} \leq t$. Now consider the numbers $pt , pt + 1$. Clearly $t < q < p$, so the largest prime factor of $pt$ is $p$. Also, since $\frac{pt + 1}{2q} \leq t < q$, the largest prime factor of $pt + 1$ is $q$. Thus we are done in this case. Let $t$ be even. Then, $r = q - t$ is odd. We have $r = q - t \equiv -t \equiv \frac{1}{p} \pmod{q}$. So we have $q \mid pr - 1$. Now we proceed similarly. Since $r$ is odd, we have $pr - 1$ is even, and so $\frac{pr - 1}{2q}$ is an integer. So, $\frac{pr - 1}{2q} < \frac{2qr - 1}{2q} = r - \frac{1}{2q}$, implying that $\frac{pr - 1}{2q} \leq r - 1 \leq q - 2$. Now we consider the numbers $pr - 1 , pr$. Clearly since $r < q < p$, largest prime divisor of $pr$ is $p$. Also , since $\frac{pr - 1}{2q} \leq q - 2$, we have that the largest prime divisor of $pr - 1$ is $q$. We are done in this case as well.