If you're not into "hunting for primes", here is a slight and concrete variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example:
$$\text{gcd}(a+i,b-i)=\text{gcd}((a-2)+i+2,b+2-(i+2))>1.$$
In fact, for $0\leq i\leq k\geq2$, we have
$$\text{gcd}(a+i,b-i)=i+2.$$

BTW, very cool drawing, Joseph!