Prove that at least $15$ students come from the same country.

The problem you've given is actually combinatorics, not number theory. A possible solution is as follows:

If there are at least $5$ countries that have at least $2$ students, we can simply take these students and obtain a contradiction. Therefore, only four countries can have at least $2$ students, and we let the set of such countries be $x$. The rest have $1$, and we let the set of such countries be $y$.

If $2|x|+|y| \ge 10$, we simply take two students from each country in $x$, and one from each in $y$, and we obtain $10$ students that fail the conditions. So $2x + y \le 10$. Note, however, that the number of students in $x$ is $60 - |y|$, and so by the pigeonhole principle, one of the countries in $x$ has $\lceil \frac{60-|y|}{|x|} \rceil \ge \lceil \frac{60-(10 - 2|x|)}{|x|} \rceil = \lceil \frac{50}{|x|} + 2 \rceil$ students. Note that $|x| \le 4$ immediately implies this is at least $15$, and we are done.

Suppose for a contradiction that there are no $15$ students from the same country. Then, if we define a set $A_i$ to be the set of students coming from country $i$, we have $|A_i| \le 14$ and $\sum_{i = 1}^n|A_i| = 60$ where $n$ is the number of countries in the conference. Now, notice that we can have at most $4$ countries with $14$ participants and in this case, we can choose $2$ participants from each and choose $2$ more from the rest of $4$ participants, which is contradictory to the assumption that for every $10$ students, there were $3$ from the same country. And if we decrease the number of participants from $14$ to something else, this will be the case still (since we still have at least $5$ countries where we can choose $2$ students) therefore we got a contradiction.