China Southeastern Mathematical Olympiad

Take $A = \{1, 2, 3, ..., 35\}$, then for any $a, b \in A$,
$$a - b \le 34,\ a + b \le 34 + 35 = 69.$$
In the following, we show that $1 \le n \le 69$. Let $A = \{a_1, a_2, \dots, a_{35}\}$, without loss of generality, suppose that $a_1 < a_2 < \dots < a_{35}$.

i.
If $1 \le n \le 19$, by
$$1 \le a_1 < a_2 < \dots < a_{35} \le 50,$$
$$2 \le a_1 + n < a_2 + n < \dots < a_{35} + n \le 50 + 19 = 69,$$
and by Dirichlet's Drawer Theorem, there exist $1 \le i, j \le 35$ ($i \ne j$) such that $a_i + n = a_j$, that is, $a_i - a_j = n$.

ii.
If $51 \le n \le 69$, by
$$1 \le a_1 < a_2 < \dots < a_{35} \le 50,$$
$$1 \le n - a_{35} < n - a_{34} < \dots < n - a_1 \le 68,$$
and by Dirichlet's Drawer Theorem, there exist at least $1 \le i, j \le 35$ ($i \ne j$) such that $n - a_i = a_j$, that is, $a_i + a_j = n$.

iii.
If $20 \le n \le 24$, since
$$50 - (2n + 1) + 1 = 50 - 2n \le 50 - 40 = 10,$$
we see that there are at least 25 elements in $a_1, a_2, \dots, a_{35}$ that belong to $[1, 2n]$.
There are at most 24 elements in $\{1, n+1\}, \{2, n+2\}, \dots, \{n, 2n\}$ such that $\{a_i, a_j\} = \{i, n+i\}$. Hence, $a_j - a_i = n$.

iv. If $25 \le n \le 34$, since $\{1, n+1\}, \{2, n+2\}, \dots, \{n, 2n\}$ have at most 34 elements, by Dirichlet Drawer Theorem, there exist $1 \le i, j \le 35$ ($i \ne j$) such that $a_i = i, a_j = n+i$, that is $a_j - a_i = n$.

v. If $n = 35$, there are 33 elements $\{1, 34\}, \{2, 33\}, \dots, \{17, 18\}, \{35\}, \{36\}, \dots, \{50\}$. Hence, there exist $1 \le i, j \le 35$ ($i \ne j$) such that $a_i + a_j = 35$.

vi.
If $36 \le n \le 50$,
if $n = 2k + 1, \{1, 2k\}, \{2, 2k-1\}, \dots, \{k, k+1\}, \{2k+1\}, \dots, \{50\}$;
if $18 \le k \le 20, 50 - (2k + 1) + 1 = 50 - 2k \le 50 - 36 = 14$;
if $21 \le k \le 24$, $50 - (2k + 1) + 1 = 50 - 2k \le 50 - 42 = 8$,
there exist $1 \le i, j \le 35$ ($i \neq j$) such that $a_i + a_j = 2k + 1 = n$.
If $n = 2k$, $\{1, 2k-1\}$, $\{2, 2k-2\}$, ..., $\{k-1, k+1\}$, $\{k\}$, $\{2k\}$, $\{2k+1\}$, ..., $\{50\}$;
if $18 \le k \le 19$, $50 - (2k + 1) + 3 \le 16k - 1 \le 19 - 1 = 18$;
if $20 \le k \le 23$, $50 - (2k + 1) + 3 \le 50 - 2k + 2 \le 12k - 1$
$\le 23 - 1 = 22$;
if $24 \le k \le 25$, $50 - (2k + 1) + 3 \le 50 - 2k + 2 \le 4k - 1 \le$
$25 - 1 = 24$,
there exist $1 \le i, j \le 35$ ($i \neq j$) such that $a_i + a_j = 2k$.
$\square$