Let $\phi(n,m), m \neq 1$, be the number of positive integers less than or equal to $n$ that are coprime with $m.$ Clearly, $\phi(m,m) = \phi(m)$, where $\phi(m)$ is Euler’s phi function. Find all integers $m$ that satisfy the following inequality:
\[\frac{\phi(n,m)}{n} \geq \frac{\phi(m)}{m}\]
for every positive integer $n.$
1. **Restate the problem and definitions:**
   We need to find all integers \( m \) such that the inequality
   \[
   \frac{\phi(n,m)}{n} \geq \frac{\phi(m)}{m}
   \]
   holds for every positive integer \( n \). Here, \(\phi(n,m)\) denotes the number of positive integers less than or equal to \( n \) that are coprime with \( m \), and \(\phi(m)\) is Euler’s totient function.

2. **Case 1: \( m \) is a power of a prime:**
   Let \( m = p^a \) where \( p \) is a prime and \( a \geq 1 \). We need to show that the inequality holds for every \( n \).

   - For \( m = p^a \), the function \(\phi(m)\) is given by:
     \[
     \phi(p^a) = p^a \left(1 - \frac{1}{p}\right) = p^a - p^{a-1}
     \]
   - For any \( n \), \(\phi(n, p^a)\) counts the numbers up to \( n \) that are not divisible by \( p \). This is:
     \[
     \phi(n, p^a) = n - \left\lfloor \frac{n}{p} \right\rfloor
     \]
   - Therefore, we have:
     \[
     \frac{\phi(n, p^a)}{n} = \frac{n - \left\lfloor \frac{n}{p} \right\rfloor}{n}
     \]
   - Since \(\left\lfloor \frac{n}{p} \right\rfloor \leq \frac{n}{p}\), it follows that:
     \[
     \frac{\phi(n, p^a)}{n} \geq \frac{n - \frac{n}{p}}{n} = 1 - \frac{1}{p}
     \]
   - We know that:
     \[
     \frac{\phi(p^a)}{p^a} = 1 - \frac{1}{p}
     \]
   - Thus:
     \[
     \frac{\phi(n, p^a)}{n} \geq \frac{\phi(p^a)}{p^a}
     \]
   - Hence, the inequality holds for \( m = p^a \).

3. **Case 2: \( m \) has more than one distinct prime factor:**
   Suppose \( m \) has more than one distinct prime factor, say \( m = p_1 p_2 \cdots p_r \) with \( r > 1 \).

   - Consider \( n = p_1 p_2 \cdots p_r - p_1 \). We need to evaluate \(\phi(n, m)\) and compare it with \(\phi(m)\).
   - The number of integers up to \( n \) that are coprime with \( m \) is:
     \[
     \phi(n, m) = \phi(n, p_1 p_2 \cdots p_r) = \phi(p_1 p_2 \cdots p_r) - (p_1 - 1)
     \]
     because \( p_1 p_2 \cdots p_r - i \) for \( 0 < i < p_1 \) are all relatively prime to \( m \).
   - We have:
     \[
     \phi(p_1 p_2 \cdots p_r) = p_1 p_2 \cdots p_r \left(1 - \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_r}\right)
     \]
   - Therefore:
     \[
     \frac{\phi(m)}{m} = \frac{\phi(p_1 p_2 \cdots p_r)}{p_1 p_2 \cdots p_r}
     \]
   - We need to show:
     \[
     \frac{\phi(n, m)}{n} = \frac{\phi(p_1 p_2 \cdots p_r) - (p_1 - 1)}{p_1 p_2 \cdots p_r - p_1} < \frac{\phi(p_1 p_2 \cdots p_r)}{p_1 p_2 \cdots p_r}
     \]
   - This inequality simplifies to:
     \[
     \frac{\phi(p_1 p_2 \cdots p_r) - (p_1 - 1)}{p_1 p_2 \cdots p_r - p_1} < \frac{\phi(p_1 p_2 \cdots p_r)}{p_1 p_2 \cdots p_r}
     \]
   - Which is equivalent to:
     \[
     -(p_1 - 1)p_1 p_2 \cdots p_r < -p_1 \phi(p_1 p_2 \cdots p_r)
     \]
   - Simplifying further:
     \[
     p_2 \cdots p_r > (p_2 - 1) \cdots (p_r - 1)
     \]
   - This inequality is clearly true since each \( p_i > 1 \).

4. **Conclusion:**
   The inequality holds for every positive integer \( n \) if and only if \( m \) is a power of a prime.

The final answer is \( \boxed{ m } \) is a power of a prime.