Find all positive integer $n$, such that 
$$\left[\frac{n}{2^0}\right]\left[\frac{n}{2^1}\right]\ldots\left[\frac{n}{2^k}\right]+2\cdot 4^{[\frac{k}{2}]}$$
is a square, where $k$ is the non-negative integer satisfying $2^k\leq n<2^{k+1}$. 
To solve the problem, we need to find all positive integers \( n \) such that the expression 
\[ \left[\frac{n}{2^0}\right]\left[\frac{n}{2^1}\right]\ldots\left[\frac{n}{2^k}\right] + 2 \cdot 4^{\left\lfloor \frac{k}{2} \right\rfloor} \]
is a perfect square, where \( k \) is the non-negative integer satisfying \( 2^k \leq n < 2^{k+1} \).

1. **Binary Representation and Initial Setup:**
   Given \( 2^k \leq n < 2^{k+1} \), we can express \( n \) in binary as \( n = \overline{a_k a_{k-1} \cdots a_1 a_0} \). For simplicity, we omit the overline when referring to binary numbers.

2. **Expression for \( A \):**
   Let 
   \[ A = \left\lfloor \frac{n}{2^0} \right\rfloor \left\lfloor \frac{n}{2^1} \right\rfloor \cdots \left\lfloor \frac{n}{2^k} \right\rfloor. \]
   We need \( A + 2 \cdot 4^{\left\lfloor \frac{k}{2} \right\rfloor} = t^2 \) for some integer \( t \).

3. **Prime Divisors and Modulo Conditions:**
   If \( p \) is an odd prime dividing \( A \), then 
   \[ \left( \frac{2}{p} \right) = 1, \]
   implying \( p \equiv \pm 1 \pmod{8} \). Therefore, 
   \[ \left\lfloor \frac{n}{2^j} \right\rfloor \not\equiv 3, 5 \pmod{8} \]
   for \( j = 0, 1, \ldots, k \).

4. **Binary Digit Constraints:**
   Since \( \left\lfloor \frac{n}{2^j} \right\rfloor \equiv a_{j+2} a_{j+1} a_j \pmod{8} \), consecutive 3 digits cannot be \( 011 \) or \( 101 \).

5. **Characterizing \( n \):**
   We deduce that \( a_{k+1} = 0 \) and \( a_k = 1 \). If \( a_{k-1} = 1 \), then \( a_{k+1} a_k a_{k-1} = 011 \), which is not allowed. Thus, \( a_{k-1} = 0 \). Similarly, \( a_{k-2} = 0 \) to avoid \( 101 \).

   Repeating this argument, we find that \( n \) must be of the form:
   \[ n = 1 \underbrace{00 \cdots 0}_{b_1} 1 \underbrace{00 \cdots 0}_{b_2} 1 \cdots 1 \underbrace{00 \cdots 0}_{b_r} 1 \underbrace{00 \cdots 0}_{c}, \]
   where \( b_1, b_2, \ldots, b_r \geq 2 \) and \( c \geq 0 \).

6. **Case Analysis:**
   - **Case 1: \( r = 0 \) (Power of 2):**
     If \( n = 2^c \), then \( c = k \) and 
     \[ t^2 = A + 2 \cdot 4^{\left\lfloor \frac{k}{2} \right\rfloor} = 2^k \cdot 2^{k-1} \cdot \cdots \cdot 2^0 + 2^{2 \left\lfloor \frac{k}{2} \right\rfloor + 1}. \]
     For \( k = 1 \), \( t = 2 \), \( n = 2 \).
     For \( k = 2 \), \( t = 4 \), \( n = 4 \).
     For \( k \geq 3 \), \( \frac{k(k+1)}{2} > k+1 \geq 2 \left\lfloor \frac{k}{2} \right\rfloor + 1 \), so \( \nu_2(\text{RHS}) = 2 \left\lfloor \frac{k}{2} \right\rfloor + 1 \) is odd, which is a contradiction.

   - **Case 2: \( r \geq 1 \):**
     Here, \( b_1 \geq 3 \) because if \( b_1 = 2 \), then \( \left\lfloor \frac{n}{2^{k-3}} \right\rfloor = 1001 = 9 = 3^2 \), which contradicts \( p \equiv \pm 1 \pmod{8} \).

     Using inequalities, we get:
     \[
     \nu_2(A) = \frac{b_1(b_1+1)}{2} + \frac{b_2(b_2+1)}{2} + \cdots + \frac{b_r(b_r+1)}{2} + \frac{c(c+1)}{2} \geq (b_1+1) + (b_2+1) + \cdots + (b_r+1) + (c+1) + 1 = (k+1) + 1 > 2 \left\lfloor \frac{k}{2} \right\rfloor + 1.
     \]
     Thus, 
     \[
     \nu_2(t^2) = \nu_2(A + 2^{2 \left\lfloor \frac{k}{2} \right\rfloor + 1}) = 2 \left\lfloor \frac{k}{2} \right\rfloor + 1,
     \]
     which is again impossible.

In conclusion, the only possible solutions are \( n = 2 \) and \( n = 4 \).

The final answer is \( \boxed{n = 2, 4} \).