Derive the insect model and restore its zero equilibrium

$1/[N(1-kN)]=1/N+k/(1-kN)$. Integrating gives $\ln|N|-\ln|1-kN|=t+C$, so $N/(1-kN)=Be^t$. Rearrangement gives $N=Be^t/(1+kBe^t)=1/(k+Ae^{-t})$, where $A=1/B$. This is a derivation, not substitution of the proposed answer.

Division by $N(1-kN)$ excluded $N=0$ and $N=1/k$. The latter is included by $A=0$ in the displayed family; the zero equilibrium must be stated separately. Nonconstant members are used only on intervals where their denominator is nonzero. The positive initial population here selects a member approaching $1/k$.

$1/k=10000$ gives $k=0.0001$. From $250=1/(k+A)$, $A=0.004-0.0001=0.0039$. Thus $N=1/(0.0001+0.0039e^{-t})$.

At $N=750$, $e^{-t}=(1/750-0.0001)/0.0039=37/117$. Thus $t=-\ln(37/117)\approx1.1513$ weeks, or approximately $8.059$ days, which rounds to 8 days. The question requests the nearest day, not the first integer day after the threshold.