Let's nail this. Take primes $N\ll p_1<\cdots<p_N$. Set $a\equiv -i\pmod{p_i}$ for all $i$, ensuring $p_i\mid a+i$ for all $1\le i\le N$. We now choose $b$ such that $p_i\mid b+i$ for all $i$ and $b\equiv 1\pmod{a}$. To that end, set $b=\ell a +1$ where $\ell$ satisfies: \[ \ell\equiv \frac1i +1\pmod{p_i},1\le i\le N. \]Note that as $p_1\gg N$, $1/i$ is well-defined modulo $p_i$ for all $i$. Furthermore, an $\ell$ as above exists due to CRT and that $\ell a + 1 \equiv -i-1+1\equiv -i\pmod{p_i}$, so $p_i\mid b+i$ as requested.

Let $a=(n+1)!+1$ and $b=2* (n+1)!+1$ , $(a,b)=1$ and $(a+k,b+k)= ((n+1)!+k+1,2*(n+1)!+k+1)=((n+1)!+k+1, k+1)=k+1$