Are you sure about this condition? $$ b_{107} \le b_{106} \ge \dots \ge b_{1} > 0 $$

What is $N$ ?

I guess it should be $ \sum\limits_{k=1}^{107}{b_{k}} \le N $.

We define $A_n$ as the arithmetic mean of the first $m$ of \[\dfrac{a_1}{b_1},\dfrac{a_2}{b_2},\cdots,\dfrac{a_n}{b_n}\]and define \[c_n=\dfrac{a_n}{b_n}\]Then, we have the following lemma: Lemma. $A_n\geq c_{n+1}$ Proof. We note that \[c_n=\dfrac{a_n}{b_n}\geq\dfrac{a_{n+1}}{b_n}\geq\dfrac{a_{n+1}}{b_{n+1}}=c_{n+1}\]Thus, we see that $\{c_n\}$ is a decreasing sequence. Thus, we get that if $i>j$, then $c_i\leq c_j$. In particular, this means that \[c_{n+1}=\dfrac{c_{n+1}+c_{n+1}+\cdots+c_{n+1}}{n}\leq\dfrac{c_1+c_2+\cdots+c_n}{n}=A_n\]thus completing the proof of the lemma. $\blacksquare$ We have the following corollary: Corollary. $\{A_n\}$ is a decreasing sequence. Proof. This is true as \[A_{n+1}=\dfrac{c_1+c_2+\cdots+c_{n+1}}{n+1}\leq\dfrac{c_1+c_2+\cdots+c_n+A_n}{n+1}=\dfrac{(n+1)A_n}{n+1}=A_n\] Now, if we want to show that $A_n\geq\dfrac MN$, it suffices to show $A_{107}\geq\dfrac MN$. We have a second lemma: Lemma. The inequality \[\dfrac{1}{n}\sum_{k=1}^n\dfrac{a_k}{b_k}\geq\dfrac{\sum\limits_{k=1}^na_k}{\sum\limits_{k=1}^nb_k}\]Proof. We shall proceed by induction: We use Chebyshev's Inequality for $a_1,a_2,\ldots,a_n$ and $\tfrac{1}{b_1},\dfrac{1}{b_2},\ldots,\dfrac{1}{b_n}$ to get \[n\sum_{k=1}^n\dfrac{a_k}{b_k}\geq\left(\sum_{k=1}^na_k\right)\left(\sum_{k=1}^n\dfrac{1}{b_k}\right)\]However, by the AM-HM Inequality, we have that \[\sum_{k=1}^n\dfrac{1}{b_k}\geq\dfrac{n^2}{\sum\limits_{k=1}^n b_k}\]Combining these two give us our desired proof. $\blacksquare$ Taking $n=107$ in the Lemma gives us \[A_k\geq A_{107}\geq\dfrac{\sum\limits_{k=1}^{107} a_k}{\sum\limits_{k=1}^{107} b_k}\geq\dfrac MN\]completing the proof.

Remark

Nice,above

We can prove a more general result that $$ \sum_{i=1}^{n} \frac{a_i}{b_i} \ge n\cdot \frac{\sum_{i=1}^{n} a_i}{\sum_{i=1}^{n} b_i}$$for any sequence of positive real numbers $\{a_i \}$ and $\{b_i \}$ where the sequences are inversely sorted with respect to each other. The proof uses a very similar technique to Cauchy Induction.