Get $p \mid a k+b l$.

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Get $p \mid a k+b l$.
Let $(b, p-a)=d$, i.e., $b=k d, p-a=l d, (k, l)=1$. Therefore, $a k+b l=$ $\frac{a b}{d}+\frac{(p-a) b}{d}=p k$. Hence, $p \mid a k+b l$.

We will prove that the prime number $p$ satisfying the condition can only be 11.
It is not difficult to see that when $p=5,7$, $p \nmid N_{p-4}$, so let $p \geqslant 11$. Then $\frac{N_{p-4}}{D_{p-4}}=1+\frac{1}{2}+\frac{1}{3}+$ $\left(\frac{1}{4}+\frac{1}{p-4}\right)+\left(\frac{1}{5}+\frac{1}{p-5}\right)+\cdots+\left(\frac{1}{\frac{p-1}{2}}+\frac{1}{\frac{p+1}{2}}\right)=\frac{11}{6}+p\left(\frac{1}{4(p-4)}+\right.$ $\left.\frac{1}{5(p-5)}+\cdots+\frac{1}{p^{2}-1}\right)$, clearly, $p \mid N_{p-4}$ if and only if $p \mid 11$, i.e., $p=11$.