For a positive integer $m$ we denote by $\tau (m)$ the number of its positive divisors, and by $\sigma (m)$ their sum. Determine all positive integers $n$ for which $n \sqrt{ \tau (n) }\le  \sigma(n)$
1. **Express \( n \) in terms of its prime factorization:**
   Let \( n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} \). 

2. **Define the functions \( \tau(n) \) and \( \sigma(n) \):**
   - The number of divisors of \( n \) is given by:
     \[
     \tau(n) = (\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_m + 1)
     \]
   - The sum of the divisors of \( n \) is given by:
     \[
     \sigma(n) = (1 + p_1 + p_1^2 + \cdots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{\alpha_2}) \cdots (1 + p_m + p_m^2 + \cdots + p_m^{\alpha_m})
     \]

3. **Rewrite the inequality \( n \sqrt{\tau(n)} \leq \sigma(n) \):**
   \[
   n \sqrt{\tau(n)} \leq \sigma(n)
   \]
   Substituting the expressions for \( n \), \( \tau(n) \), and \( \sigma(n) \):
   \[
   p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} \sqrt{(\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_m + 1)} \leq (1 + p_1 + p_1^2 + \cdots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{\alpha_2}) \cdots (1 + p_m + p_m^2 + \cdots + p_m^{\alpha_m})
   \]

4. **Simplify the inequality for each prime factor \( p_i \):**
   Consider the inequality for each prime factor \( p_i \):
   \[
   p_i^{\alpha_i} \sqrt{\alpha_i + 1} \leq 1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i}
   \]
   The sum of the divisors of \( p_i^{\alpha_i} \) can be written as:
   \[
   1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i} = \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}
   \]
   Therefore, the inequality becomes:
   \[
   p_i^{\alpha_i} \sqrt{\alpha_i + 1} \leq \frac{p_i^{\alpha_i + 1} - 1}{p_i - 1}
   \]

5. **Analyze the inequality for different values of \( p_i \):**
   - For \( p_i = 2 \):
     \[
     2^{\alpha_i} \sqrt{\alpha_i + 1} \leq \frac{2^{\alpha_i + 1} - 1}{2 - 1} = 2^{\alpha_i + 1} - 1
     \]
     Simplifying, we get:
     \[
     2^{\alpha_i} \sqrt{\alpha_i + 1} \leq 2^{\alpha_i + 1} - 1
     \]
     This inequality holds for small values of \( \alpha_i \).

   - For \( p_i = 3 \):
     \[
     3^{\alpha_i} \sqrt{\alpha_i + 1} \leq \frac{3^{\alpha_i + 1} - 1}{3 - 1} = \frac{3^{\alpha_i + 1} - 1}{2}
     \]
     Simplifying, we get:
     \[
     3^{\alpha_i} \sqrt{\alpha_i + 1} \leq \frac{3^{\alpha_i + 1} - 1}{2}
     \]
     This inequality holds for small values of \( \alpha_i \).

   - For larger primes \( p_i \geq 5 \), the right-hand side of the inequality becomes significantly larger than the left-hand side, making the inequality hold for most values of \( \alpha_i \).

6. **Check specific cases:**
   - For \( n = 1 \):
     \[
     1 \sqrt{1} \leq 1 \quad \text{(True)}
     \]
   - For \( n = 2 \):
     \[
     2 \sqrt{2} \leq 3 \quad \text{(True)}
     \]
   - For \( n = 4 \):
     \[
     4 \sqrt{3} \leq 7 \quad \text{(True)}
     \]
   - For \( n = 6 \):
     \[
     6 \sqrt{4} \leq 12 \quad \text{(True)}
     \]
   - For \( n = 12 \):
     \[
     12 \sqrt{6} \leq 28 \quad \text{(True)}
     \]

Conclusion:
The positive integers \( n \) for which \( n \sqrt{\tau(n)} \leq \sigma(n) \) are \( n = 1, 2, 4, 6, 12 \).

The final answer is \( \boxed{ n = 1, 2, 4, 6, 12 } \)