The bound is $3$. For readability, I'll change $(a_1, a_2)$ to $(x,y)$. Without loss of generality, $x \geq y$. We break into cases.

**Case 1 (the main case):** $y \geq 7/3$. In this case, there exists an integer $\ell$ such that $y \leq 4 \ell-3 < 4 \ell-1 \leq 3y$. As $r$ ranges from $(4 \ell-3)/y$ to $(4 \ell-1)/y$, the value of $rx$ increases by $2 (x/y) \geq 2$. Therefore, for some $r$ in this range, $rx$ must lie in an interval of the form $(4m-3,4m-1)$. For this $r$, we have $r \in [1,3]$ and $(rx, ry)$ of the desired form.

So, from now on, assume $y \leq 7/3$. Since this means $1 \leq y \leq 3$, if $4 \ell-3 \leq x \leq 4 \ell - 1$, we are done. So we may also assume that $4m-1 \leq x \leq 4m+1$ for some integer $m$.

**Case 2:** $y \leq 3x/(4m+1)$. In this case, take $r = (4m+1)/x$ to achieve $rx=4m+1$ and $ry \leq 3$. Note that $1 \leq r \leq 4m+1/4m-1 \leq 5/3 < 3$.

Note that, if $m \geq 2$, then $3x/(4m+1) \geq 3 (4m-1)/(4m+1) \geq 7/3$. So Cases $1$ and $2$ together cover all possible values for $y$ if $m \geq 2$. We are thus left to deal with $m=1$.

More precisely, we are left to deal with the triangle $T$ bounded by $x \geq 3$, $y \leq 7/3$ and $y \geq (3/5) x$. The vertices of $T$ are at $(3,9/5)$, $(3,7/3)$ and $(35/9, 7/3)$. It is easy to check that, for any point in this triangle, we can rescale it by a factor of $\leq 3$ to land in the square $[9,11] \times [5,7]$.

The equality is tight on $\{ 3 \} \times (9/5, 7/3)$ and on the mirror image $(9/5, 7/3) \times \{ 3 \}$, as suggested by Doug Zare.