Question: Find the minimal natural number $n$ such that there exists a natural number $m$ with
$$ \frac{61n}{200} < m < \frac{63n}{200} $$

I have tried to use the continued fraction of 62/200, but I do not know how to argue the minimality of $n$.

Is there a general method to solve this type of question?

Answer: 
Hint: rewrite it as
$$\frac{200}{63}<\frac{n}{m}<\frac{200}{61}$$
and compare the continued fractions of $\frac{200}{63}$ and $\frac{200}{61}$. These start with $[3;5]$ and $[3;3]$ respectively, so the answer is either $[3;3],[3;4],$ or $[3;5]$.

Experts in continued fractions will know whether these first iterates are over- or underestimates, and so will know immediately which of these is the answer. You and I can just try all three.
Even if we have no knowledge of using continued fractions to argue about over/underestimates and minimality, you can proceed using the idea of TonyK's hint.

Show that

* $\frac{200}{63} < \frac{n}{m} < \frac{200}{61}$
* $\Rightarrow n = 3m+a, 0 < a < m$.
* $\Rightarrow \frac{11}{63} < \frac{a}{m} < \frac{17}{61}$
* $\Rightarrow \frac{61}{17} < \frac{m}{a} < \frac{63}{11}$.
* Since $ 61/17 < 4 < 5 < 63/11$, conclude that $ m \geq 4$.
* If $m \geq 5$, then $ n = 3m + a > 15$.
* If $m = 4$, then $ a = 1, n = 13$. This is clearly the minimum $n$.

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As to how you can generalize this, try it for different sets of values, especially those that result in going a level deeper.