For which positive integers $n\geq4$ does there exist a convex $n$-gon with side lengths $1, 2, \dots, n$ (in some order) and with all of its sides tangent to the same circle?

1. **Restate the problem and assumptions:**
   We need to determine for which positive integers \( n \geq 4 \) there exists a convex \( n \)-gon with side lengths \( 1, 2, \dots, n \) (in some order) such that all its sides are tangent to the same circle (i.e., the polygon is circumscribable).

2. **Main claim:**
   If and only if we can find \( x_1, x_2, \ldots, x_n > 0 \) such that \( l_i = x_i + x_{i+1} \) (indices \(\pmod{n}\)) for all \( i \), then there exists a circumscribable polygon with side lengths \( l_1, l_2, \ldots, l_n \).

3. **Function \( f(r) \):**
   Consider the function \( f(r) = 2 \sum_{i=1}^n \arctan\left(\frac{x_i}{r}\right) \). We analyze the behavior of \( f(r) \) as \( r \) approaches 0 and \( +\infty \):
   \[
   \lim_{r \to 0^+} f(r) = n\pi \quad \text{and} \quad \lim_{r \to +\infty} f(r) = 0
   \]
   By the Intermediate Value Theorem, there exists some \( r_0 \) such that \( f(r_0) = 2\pi \). This implies the existence of a circumscribed polygon with inradius \( r_0 \) and tangency segments \( x_i \).

4. **Case \( n = 4k + 2 \):**
   If \( n = 4k + 2 \), the condition implies:
   \[
   l_1 - l_2 + l_3 - \cdots + l_{n-1} - l_n = 0
   \]
   Taking this equation modulo 2, we get:
   \[
   0 \equiv l_1 - l_2 + l_3 - \cdots + l_{n-1} - l_n \pmod{2}
   \]
   Since \( l_1, l_2, \ldots, l_n \) are the numbers \( 1, 2, \ldots, n \), their sum is:
   \[
   1 + 2 + \cdots + n = \frac{n(n+1)}{2}
   \]
   For \( n = 4k + 2 \):
   \[
   \frac{n(n+1)}{2} = \frac{(4k+2)(4k+3)}{2} = (2k+1)(4k+3) \equiv 1 \pmod{2}
   \]
   This is a contradiction, hence no such polygon exists for \( n = 4k + 2 \).

5. **Case \( n \) is odd:**
   For odd \( n \), consider the sequence \( (l_1, l_2, \ldots, l_n) = (1, 2, \ldots, n) \). Pick any \( x_1 \in (0, 1) \). By induction, we can show:
   \[
   x_k = (k-1) - (k-2) + (k-3) - \cdots + (-1)^k 1 + (-1)^{k+1} x_1
   \]
   If \( k \) is odd:
   \[
   x_k \geq \frac{k-1}{2} + x_1 > 0
   \]
   If \( k \) is even:
   \[
   x_k \geq \frac{k-2}{2} + 1 - x_1 > 0
   \]
   Thus, a circumscribable polygon exists for odd \( n \).

6. **Case \( n = 4t \):**
   For \( n = 4t \), consider the sequence:
   \[
   (l_1, l_2, \ldots, l_n) = (1, 2, 4, 3, 5, 6, 8, 7, \ldots, 4i+1, 4i+2, 4i+4, 4i+3, \ldots, 4k-3, 4k-2, 4k, 4k-1)
   \]
   Pick \( x_1 \in (0, 1) \). By induction:
   \[
   x_{4i+1} = x_{4i-3} = x_{4i-7} = \cdots = x_1 > 0
   \]
   \[
   x_{4i+2} = (4i+1) - x_1 > 4i \geq 0
   \]
   \[
   x_{4i+3} = (4i+2) - (4i+1) + x_1 = 1 + x_1 > 0
   \]
   \[
   x_{4i+4} = (4i+4) - (4i+2) + (4i+1) - x_1 = 4i+3 - x_1 > 0
   \]
   Thus, a circumscribable polygon exists for \( n = 4t \).

Therefore, the answer is all \( n \) not of the form \( 4k + 2 \) for some positive integer \( k \).

The final answer is \( \boxed{ n } \) not of the form \( 4k + 2 \) for some positive integer \( k \).