For each positive integer $n$ denote:
\[n!=1\cdot 2\cdot 3\dots n\]
Find all positive integers $n$ for which $1!+2!+3!+\cdots+n!$ is a perfect square.
1. **Define the sum of factorials:**
   Let \( a(n) = 1! + 2! + 3! + \cdots + n! \).

2. **Check the modulo condition for \( n \geq 4 \):**
   For \( n \geq 4 \), we need to consider the behavior of factorials modulo 5. Note that:
   \[
   1! = 1 \equiv 1 \pmod{5}
   \]
   \[
   2! = 2 \equiv 2 \pmod{5}
   \]
   \[
   3! = 6 \equiv 1 \pmod{5}
   \]
   \[
   4! = 24 \equiv 4 \pmod{5}
   \]
   For \( k \geq 5 \), \( k! \) will always be divisible by 5, hence \( k! \equiv 0 \pmod{5} \).

3. **Sum the factorials modulo 5 for \( n \geq 4 \):**
   \[
   a(n) \equiv 1! + 2! + 3! + 4! \pmod{5}
   \]
   \[
   a(n) \equiv 1 + 2 + 1 + 4 \pmod{5}
   \]
   \[
   a(n) \equiv 8 \equiv 3 \pmod{5}
   \]
   Since \( a(n) \equiv 3 \pmod{5} \) for \( n \geq 4 \), \( a(n) \) cannot be a perfect square because no perfect square is congruent to 3 modulo 5.

4. **Check small values of \( n \):**
   - For \( n = 1 \):
     \[
     a(1) = 1! = 1
     \]
     \( 1 \) is a perfect square.
   - For \( n = 2 \):
     \[
     a(2) = 1! + 2! = 1 + 2 = 3
     \]
     \( 3 \) is not a perfect square.
   - For \( n = 3 \):
     \[
     a(3) = 1! + 2! + 3! = 1 + 2 + 6 = 9
     \]
     \( 9 \) is a perfect square.

5. **Conclusion:**
   The only values of \( n \) for which \( a(n) \) is a perfect square are \( n = 1 \) and \( n = 3 \).

The final answer is \( \boxed{ n = 1 } \) and \( n = 3 \).