Let $m > 1$ be an integer. We are tasked with finding the smallest positive integer $n$ such that for any integers $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$, there exist integers $x_1, x_2, \ldots, x_n$ satisfying the following two conditions:

1. There exists $i \in \{1, 2, \ldots, n\}$ such that $x_i$ and $m$ are coprime.
2. $\sum_{i=1}^n a_i x_i \equiv \sum_{i=1}^n b_i x_i \equiv 0 \pmod{m}$.

To solve this, we need to determine the structure of $m$. Let $m = p_1^{k_1} p_2^{k_2} \cdots p_t^{k_t}$, where $p_1, p_2, \ldots, p_t$ are distinct prime factors of $m$ and $\omega(m) = t$ is the number of distinct primes dividing $m$.

We will show that the smallest $n$ satisfying the conditions is $2\omega(m) + 1$.

### Construction for $2\omega(m)$:
Consider $n = 2\omega(m)$. Write $m = p_1^{k_1} p_2^{k_2} \cdots p_t^{k_t}$ and let $\omega(m) = t$. For each $s = 1, \ldots, t$, let $p_s$ divide all of the $a_i$ and $b_i$'s except for $a_{2s-1}$ and $b_{2s}$. Consequently, $x_{2s-1}$ and $x_{2s}$ must both be divisible by $p_s$, so none of the $x_i$'s are coprime to $m$. Thus, $n = 2\omega(m)$ is not sufficient.

### Proof for $2\omega(m) + 1$:
To prove that $n = 2\omega(m) + 1$ is sufficient, we use the following claim:

Claim: For a prime $p$ and a positive integer $k$, $n$ is $p$-friendly if and only if $n$ is $p^k$-friendly.

Proof: (Sketch)
1. Base Case: For $k = 1$, the claim is trivially true.
2. Inductive Step: Assume the claim holds for $k-1$. We need to show it holds for $k$. We consider four cases based on the linear independence and zero properties of the vectors $c$ and $d$ modulo $p$. By carefully constructing $x_i$ and using the properties of linear combinations and modular arithmetic, we show that the conditions hold for $p^k$.

By the Chinese Remainder Theorem (CRT), we can reduce the problem to considering each prime power $p^k$ dividing $m$. Since the matrix formed by $a_i$ and $b_i$ has at most $2t$ entries that do not work, there exists at least one $x_i$ that is coprime to $m$ when $n = 2\omega(m) + 1$.

Therefore, the smallest positive integer $n$ satisfying the conditions is:
$$n = 2\omega(m) + 1.$$

The answer is: $\boxed{2\omega(m) + 1}$.