Given an integer $n > 1$, we aim to find the smallest positive number $m$ satisfying the following conditions: for any set $\{a, b\} \subset \{1, 2, \ldots, 2n-1\}$, there exist non-negative integers $x$ and $y$ (not both zero) such that $2n \mid ax + by$ and $x + y \leq m$.

To determine the smallest $m$, we analyze the conditions:

1. Consider $a = 1$ and $b = 2$. If $2n \mid ax + by$, then:
$$2n \leq x + 2y \leq 2(x + y) \leq 2m.$$
This implies $m \geq n$.

2. We now show that $m \leq n$.

- Case 1: If $\gcd(a, 2n) > 1$ or $\gcd(b, 2n) > 1$. Without loss of generality, assume $\gcd(a, 2n) > 1$. Choose $x = \frac{2n}{\gcd(a, 2n)}$ and $y = 0$. Then:
$$x + y = \frac{2n}{\gcd(a, 2n)} \leq \frac{2n}{2} = n.$$

- Case 2: If $\gcd(a, 2n) = 1$ and $\gcd(b, 2n) = 1$. Let $c \in [0, 2n-1]$ such that $c \equiv ba^{-1} \pmod{2n}$. The equation $ax + by \equiv 0 \pmod{2n}$ is equivalent to $x + cy \equiv 0 \pmod{2n}$. Choose $y = \left\lfloor \frac{2n}{c} \right\rfloor$ and $x = 2n - c \left\lfloor \frac{2n}{c} \right\rfloor$.

- Subcase 2.1: If $2 < c < n$, then:
$$x + y \leq c - 1 + \left\lfloor \frac{2n}{c} \right\rfloor < c - 1 + \frac{2n}{c} = \frac{(c-2)(c-n)}{c} + n + 1 < n + 1.$$
Hence, $x + y \leq n$.

- Subcase 2.2: If $c \geq n + 1$, then:
$$x + y = 2n - (c-1) \left\lfloor \frac{2n}{c} \right\rfloor = 2n - (c-1) \leq 2n - (n+1-1) = n.$$
Hence, $x + y \leq n$.

In conclusion, the smallest positive number $m$ that satisfies the given conditions is $n$. The answer is $\boxed{n}$.