How many positive integers divide the number $10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10$ ?
To determine the number of positive integers that divide \(10!\), we need to find the total number of divisors of \(10!\). This can be done by using the prime factorization of \(10!\).

1. **Prime Factorization of \(10!\)**:
   \[
   10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10
   \]
   We can rewrite each number in terms of its prime factors:
   \[
   10! = 2^8 \times 3^4 \times 5^2 \times 7^1
   \]

2. **Finding the Exponents of Prime Factors**:
   - For \(2\):
     \[
     2: \left\lfloor \frac{10}{2} \right\rfloor + \left\lfloor \frac{10}{4} \right\rfloor + \left\lfloor \frac{10}{8} \right\rfloor = 5 + 2 + 1 = 8
     \]
   - For \(3\):
     \[
     3: \left\lfloor \frac{10}{3} \right\rfloor + \left\lfloor \frac{10}{9} \right\rfloor = 3 + 1 = 4
     \]
   - For \(5\):
     \[
     5: \left\lfloor \frac{10}{5} \right\rfloor = 2
     \]
   - For \(7\):
     \[
     7: \left\lfloor \frac{10}{7} \right\rfloor = 1
     \]

3. **Total Number of Divisors**:
   The total number of divisors of a number \(n = p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}\) is given by:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   Applying this to our prime factorization:
   \[
   10! = 2^8 \times 3^4 \times 5^2 \times 7^1
   \]
   The number of divisors is:
   \[
   (8 + 1)(4 + 1)(2 + 1)(1 + 1) = 9 \times 5 \times 3 \times 2
   \]

4. **Calculating the Product**:
   \[
   9 \times 5 = 45
   \]
   \[
   45 \times 3 = 135
   \]
   \[
   135 \times 2 = 270
   \]

The final answer is \(\boxed{270}\).