Let $ n$ be a positive integer. Find all $ x_1,x_2,\ldots,x_n$ that satisfy the relation:
\[ \sqrt{x_1\minus{}1}\plus{}2\cdot \sqrt{x_2\minus{}4}\plus{}3\cdot \sqrt{x_3\minus{}9}\plus{}\cdots\plus{}n\cdot\sqrt{x_n\minus{}n^2}\equal{}\frac{1}{2}(x_1\plus{}x_2\plus{}x_3\plus{}\cdots\plus{}x_n).\]
1. We start with the given equation:
   \[
   \sqrt{x_1 - 1} + 2 \cdot \sqrt{x_2 - 4} + 3 \cdot \sqrt{x_3 - 9} + \cdots + n \cdot \sqrt{x_n - n^2} = \frac{1}{2}(x_1 + x_2 + x_3 + \cdots + x_n).
   \]

2. Let's assume \( x_i = k \cdot i^2 \) for some constant \( k \). We will determine \( k \) such that the equation holds.

3. Substitute \( x_i = k \cdot i^2 \) into the left-hand side of the equation:
   \[
   \sqrt{k \cdot 1^2 - 1} + 2 \cdot \sqrt{k \cdot 2^2 - 4} + 3 \cdot \sqrt{k \cdot 3^2 - 9} + \cdots + n \cdot \sqrt{k \cdot n^2 - n^2}.
   \]
   Simplify each term:
   \[
   \sqrt{k - 1} + 2 \cdot \sqrt{4k - 4} + 3 \cdot \sqrt{9k - 9} + \cdots + n \cdot \sqrt{n^2(k - 1)}.
   \]
   This can be written as:
   \[
   \sqrt{k - 1} + 2 \cdot \sqrt{4(k - 1)} + 3 \cdot \sqrt{9(k - 1)} + \cdots + n \cdot \sqrt{n^2(k - 1)}.
   \]
   Factor out \(\sqrt{k - 1}\):
   \[
   \sqrt{k - 1} \left( 1 + 2 \cdot 2 + 3 \cdot 3 + \cdots + n \cdot n \right).
   \]
   The sum inside the parentheses is:
   \[
   1 + 4 + 9 + \cdots + n^2 = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.
   \]
   Therefore, the left-hand side becomes:
   \[
   \sqrt{k - 1} \cdot \frac{n(n+1)(2n+1)}{6}.
   \]

4. Now, substitute \( x_i = k \cdot i^2 \) into the right-hand side of the equation:
   \[
   \frac{1}{2} \left( k \cdot 1^2 + k \cdot 2^2 + k \cdot 3^2 + \cdots + k \cdot n^2 \right).
   \]
   Factor out \( k \):
   \[
   \frac{k}{2} \left( 1^2 + 2^2 + 3^2 + \cdots + n^2 \right).
   \]
   Using the sum of squares formula again:
   \[
   \frac{k}{2} \cdot \frac{n(n+1)(2n+1)}{6}.
   \]

5. Equate the left-hand side and the right-hand side:
   \[
   \sqrt{k - 1} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{k}{2} \cdot \frac{n(n+1)(2n+1)}{6}.
   \]
   Simplify by canceling out the common terms:
   \[
   \sqrt{k - 1} = \frac{k}{2}.
   \]

6. Square both sides to solve for \( k \):
   \[
   k - 1 = \left( \frac{k}{2} \right)^2.
   \]
   This simplifies to:
   \[
   k - 1 = \frac{k^2}{4}.
   \]
   Multiply both sides by 4 to clear the fraction:
   \[
   4k - 4 = k^2.
   \]
   Rearrange into a standard quadratic equation:
   \[
   k^2 - 4k + 4 = 0.
   \]
   Factorize:
   \[
   (k - 2)^2 = 0.
   \]
   Therefore:
   \[
   k = 2.
   \]

7. Substitute \( k = 2 \) back into \( x_i = k \cdot i^2 \):
   \[
   x_i = 2 \cdot i^2.
   \]

Conclusion:
\[
\boxed{x_i = 2i^2 \text{ for } i = 1, 2, \ldots, n.}
\]