Let $n\ge 2$ be a positive integer. Find the positive integers $x$
\[\sqrt{x+\sqrt{x+\ldots +\sqrt{x}}}<n \]
for any number of radicals.
1. Let \( x > 0 \) and define the sequence \( x_m \) as follows:
   \[
   x_m = \sqrt{x + \sqrt{x + \ldots + \sqrt{x}}} \quad \text{(with \( m \) roots)}
   \]
   We need to show that the sequence \( (x_m)_{m=1}^{\infty} \) is positive and strictly increasing.

2. For \( m = 1 \):
   \[
   x_1 = \sqrt{x} > 0
   \]

3. For \( m = 2 \):
   \[
   x_2 = \sqrt{x + \sqrt{x}} > \sqrt{x} = x_1 > 0
   \]

4. Assume \( x_k > x_{k-1} > 0 \) for some \( k > 1 \). Then:
   \[
   x_{k+1} = \sqrt{x + x_k} > \sqrt{x + x_{k-1}} = x_k > 0
   \]
   By induction, the sequence \( (x_m)_{m=1}^{\infty} \) is positive and strictly increasing.

5. Next, we show that the sequence is bounded above. We start with:
   \[
   x_1 = \sqrt{x} < x + 1
   \]

6. Assume \( x_k < x + 1 \) for some \( k \ge 1 \). Then:
   \[
   x_{k+1} = \sqrt{x + x_k} < \sqrt{x + (x + 1)} = \sqrt{2x + 1} < x + 1
   \]
   By induction, \( x_m < x + 1 \) for all \( m \ge 1 \).

7. Since the sequence \( (x_m)_{m=1}^{\infty} \) is strictly increasing and bounded above, it converges to a limit \( L > 0 \).

8. Let \( L = \lim_{m \to \infty} x_m \). Then:
   \[
   L = \sqrt{x + L}
   \]
   Squaring both sides, we get:
   \[
   L^2 = x + L
   \]
   Rearranging, we obtain the quadratic equation:
   \[
   L^2 - L - x = 0
   \]

9. Solving the quadratic equation for \( L \):
   \[
   L = \frac{1 + \sqrt{1 + 4x}}{2}
   \]
   (We discard the negative root since \( L > 0 \)).

10. We need \( x_m < n \) for every \( m \ge 1 \). Since the sequence is strictly increasing and convergent, this is equivalent to:
    \[
    L \le n
    \]
    Substituting \( L \):
    \[
    \frac{1 + \sqrt{1 + 4x}}{2} \le n
    \]
    Multiplying both sides by 2:
    \[
    1 + \sqrt{1 + 4x} \le 2n
    \]
    Subtracting 1 from both sides:
    \[
    \sqrt{1 + 4x} \le 2n - 1
    \]
    Squaring both sides:
    \[
    1 + 4x \le (2n - 1)^2
    \]
    Simplifying:
    \[
    4x \le (2n - 1)^2 - 1
    \]
    \[
    4x \le 4n^2 - 4n
    \]
    Dividing by 4:
    \[
    x \le n^2 - n
    \]

11. Therefore, the positive integers \( x \) that satisfy the given inequality are:
    \[
    1 \le x \le n^2 - n
    \]

The final answer is \( \boxed{ 1 \le x \le n^2 - n } \).