Answer: $n=13224$. Let $d_k+1=a>1,2d_{k-1}+65=b$. We have $n=d_kd_{k+1}$ and $n=d_{k+2}d_{k-1}$, so $n=a^2-1$ and $4n+65^2=b^2$, so $(b-2a)(b+2a)=63 \cdot 67$. Since $67$ is prime we have $67|b+2a$, because if $67|b-2a$ then $63\geq b+2a > b-2a \geq 67$, which is false. But if $b+2a=67,b-2a=63$ then $a=1$, but $a>1$. So, since $2 \not | 63 \cdot 67$ we have $b+2a \geq 3 \cdot 67$ and so $4a=(b+2a)-(b-2a) \geq 3 \cdot 67 - 21=120$ and $a \geq 30$ and so $n \geq 899$. But $n=899$ doesn't works, so $b + 2a \geq 7 \cdot 67$ ($7$ is next divisor of $63$ after $3$) and so $a \geq 115$ and $n \geq 115^2-1=13224$. It works: we have $k=16, d_{k-1}=87,d_k=114,d_{k+1}=116,d_{k+2}=152$.

@above answer is 2024

A natural number $n$ is good If it has an even number of positive divisors (i.e. $2 \mid \tau(n)$) $(1) \;\; 1 = d_1 < d_2 < \cdots < d_{2k-1} < d_{2k}=n$ satisfying the equations $(2) \;\; d_{k+1} - d_k = 2$, $(3) \;\; d_{k+2} - d_{k-1} = 2$. The fact that $\rho(n)$ is even means $n$ is not a perfect square. Hence there is a natural number $m$ s.t. $(4) \;\; m^2 < n < (m+ 1)^2$. Moreover $(5) \;\; d_k < \sqrt{n} < d_{k+1}$, which according to inequalities (4) implies $d_k \leq m < d_{k+1}$, yielding $(d_k,d_{k+1}) = (m-1,m), (m,m+2)$. The inequalities (4) give us $(6) \;\; n = m^2 + r, 1 \leq r \leq 2m$. Assume $(d_k,d_{k+1})=(m-1,m+1)$. Then ${\textstyle \frac{n}{d_{k+1}} = \frac{m^2 + r}{m + 1} = \frac{m(m+ 1) + (r - m)}{m + 1} = m + \frac{r - m}{m + 1} \in \mathbb{N}}$. yielding $m + 1 \mid r - m$, which implies $r=m$ since $-(m - 1) \leq r - m \leq m$, i.e. $|r - m| < m+1$. Therefore $n = m^2 + m$, yielding $m \mid n$, which means $d_k=m$. This contradiction give us $(d_k,d_{k+1}) = (m,m+2)$. Applying the formula $d_i \cdot d_j= n$ when $i + j = 2k+1$ we obtain $d_k \cdot d_{k+1} = n$, i.e. $(7) \;\; n = m(m + 2)$. Likewise we know that $d_{k-1} \cdot d_{k+2} = n$, which combined with formulas (3) and (7) result in ${\textstyle (8) \;\; \frac{m^2 + m}{d_{k-1}} - d_{k-1} = 65}$. Setting $x=d_{k-1}$, equation (8) becomes $m^2 + 2m - x^2 = 65x$, or alternatively $(2x + 65)^2 - (2m + 2)^2 = 65^2 - 2^2$, i.e. $(2x + 65 - 2m - 2)(2x + 65 + 2m + 2) = (65 - 2)(65 + 2)$, which implies ${\textstyle (9) \;\; (2x + 65 - 2m - 2,2x + 65 + 2m + 2) = (d,\frac{4221}{d})}$, where $d \leq \sqrt{4221} < 65$ and $(10) \;\; d \mid 3^2 \cdot 7 \cdot 67$. According to formulas (9) ${\textstyle (11) \;\; 4(m + 1) = \frac{4221}{d} - d}$. The minimal value of $m$ are obtain by finding the largest value of $d$ less than 65, which according to condition (10) is $d = 3 \cdot 7 = 21$. Hence by formula (11) the minimal value of $m$ is ${\textstyle m = \frac{1}{4}(\frac{4221}{21} - 21) - 1 = \frac{201 - 21}{4} - 1 = \frac{180}{4} - 1 = 45 - 1 = 44}$. Hence by formula (7) the smallest good number is $n = 44 \cdot 46 = 2024$.