Aren't they the same value?

Since $\dfrac{33}{100} < \dfrac{m}{n}$ , we have $100m > 33n$ and therefore $100m \geq 33n+1$ .

Since $\dfrac{m}{n} < \dfrac{1}{3}$ , we obtain $n > 3m$ and therefore $n \geq 3m+1$ .

Hence $100m \geq 33(3m+1)+1 = 99m + 34$ and $m \geq 34$ .

Therefore $n \geq 3 \times 34 + 1 = 103.$

Since $\dfrac {33}{100} < \dfrac {34}{103} < \dfrac {1}{3}$ , the least $n$ is 103.

Taking reciprocals of everything, we have $3 < \frac{n}{m} < 3 \frac{1}{33}$ .

Multiplying both sides by $m$ we get $3m < n < 3m + \frac{m}{33}$ .

It is obvious that if we want to fit an integer $n$ between $3m$ and $3m +\frac{m}{33}$ (keeping in mind that $3m$ is an integer), then $m \geq 34$ .

Value $m=34$ yields $n=103$ (and obviously larger $m$ yields larger $n$ ).

Multiplying the inequality by $3n$ gives

$.99n<3m<n$

That means there must be an integer that is divisible by $3$ between $.99n$ and $n$ . Since $n=100$ gives $99$ and $100$ , we know that for an integer to be between $.99n$ and $n$ , $n \geq 101$ . The smallest integer that is divisible by $3$ and greater than $100$ is $102$ , which is between $.99*103$ and $103$ .

Thus, our answer is $\boxed{103}$

$\text{Based on the fact that for b>a, n>0, } \dfrac a b <\dfrac {a+n}{b+n} we~have:-\\ \dfrac 1 3 =.333333333.~\text{So we need .33< }\dfrac m n <.333333\\.33=\dfrac {33}{100}< \dfrac {33+1}{100+3}=.3300970874<\dfrac 1 3\\ \therefore~\dfrac {33+1}{100+3}=\dfrac {34}{103} ~ ~ ANSWER~~\color{#D61F06}{103},\\~~\\~~\\\text {If we had }.33< \dfrac m n <.330095,\text{ we would have gone for more} \\ \text{ refinement as follows.}\\\dfrac{33+\frac 1 2 }{100+\frac 3 2 } =.330049, ~~and...then ~\dfrac{2*(33+\frac 1 2) }{2*(100+\frac 3 2 )} \\=\dfrac{67 }{203} =.3300492611~~~~~\text{By refining fraction to be added and}\\\text {compensating. Note:- By this, we are securing our lower bound for} \\ \text{ minimum value. Fractions has to be as reciprocal of integers only. }$

Mediant fraction inequality states that : a/b<(a+c)/(b+d)<c/d