Arbitrarily long composite anti-diagonals?

Fix integer $k\ge 1$, and let $p_i$ denote the $i$th prime. By the Chinese Remainder Theorem, there exist integer $a,b>k$ such that $a\equiv -i\pmod{p_i}$ and $b\equiv i\pmod{p_i}$ for each $i=1,\dotsc,k$. Now $p_i\mid\gcd(a+i,b-i)$ for $i=1,\dotsc, k$, which yields a length-$k$ antidiagonal.

In the same way one can show, say, that for any integers $x_1,\dotsc,z_k$ there exist (arbitrarily large) integers $a,b,c$ such that $\gcd(a+x_i,b+y_i,c+z_i)>1$ for each $i\in[1,k]$.

Bit more explicit form of Seva's example is the following. Assume that both $a-2,b+2$ are divisible by all primes between 2 and $k+2$. Then $$gcd(a+i,b-i)=gcd((a-2)+i+2,b+2-(i+2))>1$$ for all $i=0,\dots,k$.

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: for $0\leq i\leq k\geq2$, we have $$\text{gcd}(a+i,b-i)=i+2>1.$$

BTW, very cool drawing, Joseph!