Let $p_1<p_2<\dots<p_m$ be the primes less than $1402$ and $p_{m+1}<p_{m+2}<\dots$ the sequence of other primes. Consider $n=AB$ where $A=(p_1p_2\dots p_m)^{1402}$ and $b=p_{m+1}\dots p_k$. We claim that this does the trick for $k$ sufficiently large. So take any $i \le 1402$ and any sign, so that we need to prove that $\tau(n)>1401\tau(n \pm i)$. First of all, note that $\tau(n) \ge 2^k$. Since $\tau(N) \le 2^{\Omega(N)}$ for general $N$, where $\Omega(N)$ is the number of prime factors with repetition (suffices to check for prime powers where it follows from $2^k \ge k+1$), it suffices to prove that $\Omega(n \pm i) \le k-12$. Note that each of the primes $p_1,\dots,p_m$ divides $n \pm i$ at most once by construction, and none of $p_{m+1},\dots,p_k$ divide $n \pm i$. Write $n \pm i=ab$ where $b$ is the part coprime to $p_1,\dots,p_k$. Hence $b \le n+i \le 2n \le Ap_1p_2\dots p_k$ and it suffices to prove that $\Omega(b) \le k-2023$. But this is rather easy: Otherwise, we would have $b \ge p_k^{k-2023}$ and hence \[p_k^{k-2023} \le Ap_1\dots p_k \le Ap_1\dots p_{2024} p_k^{k-2024}\]which is equivalent to $p_k \le Ap_1\dots p_{2024}$ and hence clearly fails for large $k$. Done!