Find all positive integer $n$ such that for all $i=1,2,\cdots,n$, $\frac{n!}{i!(n-i+1)!}$ is an integer.

[i]Proposed by ckliao914[/i]
To find all positive integers \( n \) such that for all \( i = 1, 2, \ldots, n \), the expression \(\frac{n!}{i!(n-i+1)!}\) is an integer, we proceed as follows:

1. **Expression Simplification**:
   \[
   \frac{n!}{i!(n-i+1)!}
   \]
   This can be rewritten using the binomial coefficient:
   \[
   \frac{n!}{i!(n-i+1)!} = \frac{n!}{i!(n-i)!(n-i+1)} = \frac{1}{n-i+1} \binom{n}{i}
   \]
   For this to be an integer, \(\frac{1}{n-i+1} \binom{n}{i}\) must be an integer for all \( i = 1, 2, \ldots, n \).

2. **Prime Divisors**:
   Let \( p \) be a prime divisor of \( n+1 \). Since \( p \leq n \), consider the term:
   \[
   \frac{n!}{p!(n-p+1)!} = \frac{(n-p+2)(n-p+3) \cdots n}{p!}
   \]
   For this to be an integer, \( p \) must divide the numerator \((n-p+2)(n-p+3) \cdots n\).

3. **Modulo \( p \) Analysis**:
   Consider the product \((n-p+2)(n-p+3) \cdots n\) modulo \( p \):
   \[
   (n-p+2)(n-p+3) \cdots n \equiv 1 \cdot 2 \cdot \ldots \cdot (p-1) \pmod{p}
   \]
   Since \( 1 \cdot 2 \cdot \ldots \cdot (p-1) \equiv (p-1)! \equiv -1 \pmod{p} \) by Wilson's Theorem, it is not divisible by \( p \).

4. **Special Case \( n = p-1 \)**:
   If \( n = p-1 \) for some prime \( p \), then:
   \[
   \frac{n!}{i!(n-i+1)!} = \frac{(p-1)!}{i!(p-i)!}
   \]
   This can be rewritten as:
   \[
   \frac{(p-1)!}{i!(p-i)!} = \binom{p-1}{i}
   \]
   Since \(\binom{p-1}{i}\) is always an integer, the condition is satisfied.

5. **Conclusion**:
   The only positive integers \( n \) that satisfy the condition for all \( i = 1, 2, \ldots, n \) are of the form \( n = p-1 \) where \( p \) is a prime number.

The final answer is \( \boxed{ n = p-1 } \) where \( p \) is a prime number.