Find all natural numbers $n$ such that $1+[\sqrt{2n}]~$ divides $2n$.

( For any real number $x$ , $[x]$ denotes the largest integer not exceeding $x$. )
To find all natural numbers \( n \) such that \( 1 + \lfloor \sqrt{2n} \rfloor \) divides \( 2n \), we proceed as follows:

1. Let \( x = \lfloor \sqrt{2n} \rfloor \). By definition, \( x \) is the largest integer not exceeding \( \sqrt{2n} \), so we have:
   \[
   x \leq \sqrt{2n} < x + 1
   \]
   Squaring all parts of the inequality, we get:
   \[
   x^2 \leq 2n < (x + 1)^2
   \]
   This simplifies to:
   \[
   x^2 \leq 2n < x^2 + 2x + 1
   \]

2. We are given that \( 1 + x \) divides \( 2n \). Let \( 2n = k(1 + x) \) for some integer \( k \). Then:
   \[
   2n = k(1 + x)
   \]

3. From the inequality \( x^2 \leq 2n < x^2 + 2x + 1 \), we substitute \( 2n = k(1 + x) \) to get:
   \[
   x^2 \leq k(1 + x) < x^2 + 2x + 1
   \]

4. We need to find \( k \) such that the above inequality holds. First, consider the lower bound:
   \[
   x^2 \leq k(1 + x)
   \]
   Dividing both sides by \( 1 + x \) (which is positive since \( x \) is a natural number):
   \[
   \frac{x^2}{1 + x} \leq k
   \]

5. Now consider the upper bound:
   \[
   k(1 + x) < x^2 + 2x + 1
   \]
   Dividing both sides by \( 1 + x \):
   \[
   k < \frac{x^2 + 2x + 1}{1 + x} = x + 1
   \]

6. Combining the two inequalities, we get:
   \[
   \frac{x^2}{1 + x} \leq k < x + 1
   \]

7. Since \( k \) must be an integer, the only possible value for \( k \) is \( x \) itself. Thus:
   \[
   k = x
   \]

8. Substituting \( k = x \) back into \( 2n = k(1 + x) \), we get:
   \[
   2n = x(1 + x)
   \]
   Simplifying, we find:
   \[
   2n = x^2 + x
   \]
   Therefore:
   \[
   n = \frac{x(x + 1)}{2}
   \]

9. Since \( x \) is a natural number, \( n \) must be a triangular number. The natural numbers \( n \) that satisfy the given condition are those that can be expressed as \( \frac{x(x + 1)}{2} \) for some natural number \( x \).

The final answer is \( \boxed{ n = \frac{x(x + 1)}{2} } \).