Let $n = p_1^{\alpha_1} \cdots p_s^{\alpha_s}$. We define $\Omega(n) = \alpha_1 + \cdots + \alpha_s$.
Lemma 1. $d(n) \le 2^{\Omega(n)}$
Proof.
$$\forall 1 \le i \le s : 2^{\alpha_i} \ge \alpha_i + 1 \implies 2^{\Omega(n)} \ge d(n)$$

We denote prime numbers less than $1402$ as $q_1, \dots, q_l$ and the rest of the prime numbers as $p_1 < p_2 < \dots$. We choose $n = q_1^\alpha \cdots q_l^\alpha p_1 \cdots p_k$ where $\alpha, k$ will be determined later.
Suppose $\alpha$ is large enough such that
$$\forall 1 \le t \le l,\ i \le 1402 : v_{q_t}(i) < \alpha$$
To do so we restrict $\alpha > \log_2 1402$. Now if for a prime number like $q$ and a number less than $1402$ like $i$ we have $q \mid i$ then $q \in \{q_1, \dots, q_l\}$ so $q \mid \frac{n}{i}$. Thus $\gcd(i, \frac{n}{i} \pm 1) = 1$.
$$d(n \pm i) = d(i)d\left(\frac{n}{i} \pm 1\right) \le 1402 \cdot 2^{\Omega\left(\frac{n}{i} \pm 1\right)} \le 1402 \cdot 2^{\log_{p_{k+1}}\left(\frac{n}{i} \pm 1\right)} \le 1402 \cdot 2^{\log_{p_{k+1}}(n+1)}$$
It is enough to show that
$$(\alpha + 1)^l \times 2^k > 1401 \times 1402 \times 2^{\log_{p_{k+1}}(n+1)}$$
By adding the condition to the choice of $\alpha$ that $(\alpha + 1)^l > 1401 \times 1402$ it remains to prove $k \ge \log_{p_{k+1}}(n+1)$.
$$\iff p_{k+1}^k > q_1^\alpha \cdots q_l^\alpha p_1 \cdots p_k$$
We may choose $k$ such that $p_k > q_1^\alpha \cdots q_l^\alpha$. This ends the proof.

We shall indeed prove that for all $k, m$ there is a positive integer $n > 1$ such that $\frac{d(n)}{d(n \pm i)} > m$, $i = 1, \dots, k$. Let $C = \max\{d(1), \dots, d(k)\}$ and let $s$ be a positive integer such that $2^{s-1} > Cm$. Let $p_i$ be the $i$-th prime number, chose a positive integer $l$ such that $p_{s+l} > k!p_1 \dots p_s$. Taking $n = k!p_1 \dots p_{s+l}$ then $n \pm i$ for all $i = 1, \dots, k$ is greater than $i$ and is also divisible by $i$. Then, for all $i = 1, \dots, k$;
$$n \pm i = i q_1^{\alpha_1} \dots q_r^{\alpha_r},$$
Where $q_1 < \dots < q_r$ be prime numbers. Thus,
$$n \pm i = i \left( \frac{k!}{i} p_1 \dots p_{s+l} \pm 1 \right).$$
That is, $\frac{n \pm i}{i}$ are divisible by primes greater than $p_1, \dots, p_{s+l}$. Therefore, $q_1 > p_{s+l} > k!p_1 \dots p_s$ that is, $q_1 > 1 + k!p_1 \dots p_s$ hence, $p_{k+1} \dots p_{l+s} < q_1^s$. Thus,
$$q_1^{s+1} > k! p_1 \dots p_s + q_1^s = n + q_1^s > n + q_1 \ge n + k.$$
Yielding $q_1^{l+1} > n \pm i$. That is, $q_1^{l+1} > q_1^{\alpha_1} \dots q_r^{\alpha_r} \ge q_1^{\alpha_1+\dots+\alpha_r}$.

Then, $l+1 > \alpha_1 + \dots + \alpha_r \ge r$ to find $l+1+r > (1+\alpha_1)+\dots+(1+\alpha_r)$.
Hence,
$$d(n \pm i) = d(i)(1 + \alpha_1) + \dots + (1 + \alpha_r) \le C \left( \frac{(1 + \alpha_1) + \dots + (1 + \alpha_r)}{r} \right)^r < C \left( \frac{l+1+r}{r} \right)^r$$
Finally, $(1 + \frac{l+1}{r})^r < (1 + \frac{l+1}{l+1})^{l+1} = 2^{l+1}$. Yielding $d(n \pm i) < C \cdot 2^{l+1}$. Since $d(n) > 2^{s+1}$ we find that
$$\frac{d(n)}{d(n \pm i)} > \frac{2^{s-1}}{C} > m.$$