The product $1\times 2\times  3\times ...\times  n$  is written on the board.  For what integers $n \ge 2$, we can add exclamation marks to some factors to convert them into factorials, in such a way that the final product can be a perfect square?
To determine for which integers \( n \geq 2 \) we can add exclamation marks to some factors to convert them into factorials such that the final product is a perfect square, we need to analyze the prime factorization of the product \( 1 \times 2 \times 3 \times \ldots \times n \).

1. **Prime Factorization and Perfect Squares**:
   - A number is a perfect square if and only if all the exponents in its prime factorization are even.
   - Consider the product \( n! = 1 \times 2 \times 3 \times \ldots \times n \). We need to check if we can add exclamation marks to some factors to make the product a perfect square.

2. **Prime \( n \)**:
   - If \( n \) is a prime number, then \( n! \) includes \( n \) itself, which has an exponent of 1 in the prime factorization of \( n! \). This makes it impossible to convert \( n! \) into a perfect square by adding exclamation marks to any of the factors.

3. **Composite \( n \)**:
   - For composite \( n \), we can use the following algorithm to convert \( n! \) into a perfect square:
     - Let \( S_i \) denote the number on the board after \( i \) exclamations (initially \( S_0 = n! \)).
     - Let \( p_i \) denote the largest prime such that \( v_{p_i}(S_i) \) (the exponent of \( p_i \) in the prime factorization of \( S_i \)) is odd.
     - The algorithm proceeds as follows:
       1. If \( p_i \) does not exist (i.e., all exponents are even), the algorithm terminates, and \( S_i \) is a perfect square.
       2. Otherwise, apply an exclamation mark after \( p_i + 1 \). This means we replace \( p_i + 1 \) with \( (p_i + 1)! \), which increases the exponents of the primes in \( (p_i + 1)! \).

4. **Proof of Algorithm**:
   - Each time we apply an exclamation mark to \( p_i + 1 \), the largest prime \( p_i \) with an odd exponent decreases.
   - Since the set of primes is finite and \( p_i \) decreases with each step, eventually, there will be no primes with odd exponents, making \( S_i \) a perfect square.

Therefore, the algorithm works for all composite \( n \).

The final answer is all composite \( \boxed{ n \geq 2 } \).