There are $200$ numbers on a blackboard:
$ 1! ,  2! ,  3! ,  4! , ... ... ,  199! ,  200!$.
Julia erases one of the numbers. When Julia multiplies the remaining $199$ numbers, the product is a perfect square. Which number was erased?
1. **Define the product \( N \):**
   \[
   N = 1! \cdot 2! \cdot 3! \cdot \ldots \cdot 199! \cdot 200!
   \]
   We need to determine which number, when erased, makes the product of the remaining numbers a perfect square.

2. **Factorize \( N \):**
   \[
   N = \prod_{k=1}^{200} k!
   \]
   We can split this product into two parts:
   \[
   N = \prod_{k=1}^{100} (2k-1)! \cdot (2k)!
   \]
   Notice that:
   \[
   (2k)! = (2k) \cdot (2k-1)!
   \]
   Therefore:
   \[
   N = \prod_{k=1}^{100} (2k-1)! \cdot (2k) \cdot (2k-1)!
   \]
   Simplifying further:
   \[
   N = \prod_{k=1}^{100} [(2k-1)!]^2 \cdot 2k
   \]

3. **Separate the even and odd factorials:**
   \[
   N = \left( \prod_{k=1}^{100} [(2k-1)!]^2 \right) \cdot \left( \prod_{k=1}^{100} 2k \right)
   \]
   The product of the even numbers \( 2k \) can be written as:
   \[
   \prod_{k=1}^{100} 2k = 2^{100} \cdot \prod_{k=1}^{100} k = 2^{100} \cdot 100!
   \]
   Thus:
   \[
   N = \left( \prod_{k=1}^{100} [(2k-1)!]^2 \right) \cdot 2^{100} \cdot 100!
   \]

4. **Simplify the expression:**
   \[
   N = \left( \prod_{k=1}^{100} [(2k-1)!]^2 \right) \cdot 2^{100} \cdot 100!
   \]
   To find the product of the remaining numbers when one number is erased, we divide \( N \) by the factorial of the erased number. Let’s denote the erased number by \( k! \).

5. **Determine the perfect square condition:**
   We need the product of the remaining numbers to be a perfect square. This means:
   \[
   \frac{N}{k!} \text{ must be a perfect square}
   \]
   Let’s consider the case when \( k = 100 \):
   \[
   \frac{N}{100!} = \frac{\left( \prod_{k=1}^{100} [(2k-1)!]^2 \right) \cdot 2^{100} \cdot 100!}{100!}
   \]
   Simplifying:
   \[
   \frac{N}{100!} = \left( \prod_{k=1}^{100} [(2k-1)!]^2 \right) \cdot 2^{100}
   \]
   Notice that:
   \[
   \left( \prod_{k=1}^{100} [(2k-1)!]^2 \right) \cdot 2^{100}
   \]
   is a perfect square because it is the square of:
   \[
   \left( \prod_{k=1}^{100} (2k-1)! \right) \cdot 2^{50}
   \]

6. **Conclusion:**
   Since \(\frac{N}{100!}\) is a perfect square, the number that Julia erased must be \(100!\).

The final answer is \(\boxed{100!}\).