We claim that the answer is $\boxed{730}.$ Firstly, we will show that $730$ works. WLOG let's label the students $a_1, a_2, \cdots, a_{730}$ according to their ranking in algebra, so that $A(a_1) > A(a_2)> \cdots > A(a_{730}),$ where $A(s)$ denotes the score of student $s$ in algebra. Define $G(s), C(s), N(s)$ similarly. Let the permutation $g = g_1, g_2, \cdots, g_{730}$ of $\{1, 2, \cdots, 730\}$ be such that $G(a_{g_1})> G(a_{g_2}) > \cdots > G(a_{g_{730}}).$ Notice that there cannot be an increasing subsequence of $g$ of length $9+1 = 10$, and therefore Erdos-Szekeres guarantees that there is a decreasing subsequence of length $\frac{730-1}{9} + 1 = 82.$ Let these students be $a_{i_1}, a_{i_2}, \cdots, a_{i_{82}}$ for $i_1 > i_2 > \cdots > i_{82}.$ Now, consider $C(a_{i_1}), C(a_{i_2}), \cdots, C(a_{i_{82}}).$ We know by Erdos-Szekeres that there is either an increasing subsequence of length $10$ or a decreasing subsequence of length $10$. In the case of an increasing subsequence of length $10$, those $10$ students would have the same order in both algebra and combinatorics, and so we're done. In the case of a decreasing subsequence of length $10$, those $10$ students would have the same order in both geometry and combinatorics, and so we're done again. What about number theory? Well, it turns out that the fourth subject number theory doesn't actually affect the answer at all, it just makes our life harder for the construction of $729$ students. Let's now turn to the construction of $729$ students and rankings of their abilities in the four subjects. Firstly, let's associate to each student an ordered triple $(i, j, k) \in \{1, 2, \cdots, 9\}^3.$ Now, the construction is given below: $$A(i, j, k) = 81i + 9j + k,$$$$G(i, j, k) = 81i - 9j - k,$$$$C(i, j, k) = -81x + 9y - z,$$$$N(i, j, k) = -81x -9y + z.$$In other words, $A$ ranks the students first by increasing order of $i$, then tiebreaks by increasing order of $j$, and finally tiebreaks by increasing order of $k$. Similar statements can be said about the other subjects. We will now show that these work. Suppose, for contradiction, that there were ten students $(i_1, j_1, k_1), (i_2, j_2, k_2), \cdots, (i_{10}, j_{10}, k_{10})$ which were ranked the same by two different subjects. Case 1. There are two students among the ten which have different $i$ values Suppose that $i_1 \neq i_2$, WLOG. Our two subjects must be $A$ and $G$ or $C$ and $N$, by simply looking at how the subjects rank these first two students. By Pigeonhole, there are two $1 \le a, b \le 10$ with $i_a = i_b.$ However, this is a contradiction, since $(i, j, k)$ and $(i, j', k')$ are ranked differently by $A$ and $G$, and differently by $C$ and $N$ whenever $(j, k) \neq (j', k').$ Case 2. All the $i$ values are equal, but there are two students with different $j$ values Suppose that $j_1 \neq j_2,$ WLOG. Our two subjects must be $A$ and $C$ or $G$ and $N$, by simply looking at how the subjects rank these first two students. By Pigeonhole, there are two $1 \le a, b \le 10$ with $j_a = j_b.$ However, this is a contradiction, since $(i, j, k)$ and $(i, j, k')$ are ranked differently by $A$ and $C$, and differently by $G$ and $N$ whenever $k \neq k'$. Case 3. All the $i$ values are equal, and also all $j$ values are equal This is a clear contradiction, since Pigeonhole implies that two students have the same $k$ values. This means they're the same student, clearly a contradiction. As we've obtained a contradiction in all cases, our initial assumption that there were ten students ranked the same by two different subjects was wrong. Therefore, this is a valid construction which shows that $N > 729$. Hence, we've shown that the answer is $\boxed{730},$ as claimed. $\square$

Absolutely same problem wrote: There are $N$ houses in a city. Every Christmas, Santa visits these $N$ houses in some order. Show that if $N$ is large enough, then it holds that for three consecutive years there are always are $13$ houses such that they have been visited in the same order for two years (out of the three consecutive years). Determine the smallest $N$ for which this holds. $\textbf{Wlog}, $ he visits the houses in order $1-N $ in the first year. In year two $(N>12*144), $ he has to visit an increasing sequence of $13$ houses. Which would be the same order as first year, or he has to visit a decreasing sequence of $145$ houses. In third year, note that the $145$ houses, were of decreasing order in year $2. $ He has to have an increasing subsequence of length $13, (\text {same as first year})$ or a decreasing subsequence of length $13$( as year $2) . $