To determine all sets of real numbers $S$ satisfying the given conditions, let's carefully analyze these conditions:

1. Condition 1: $1$ is the smallest element of $S$.

2. Condition 2: For all $x,y \in S$ such that $x > y$, the expression $\sqrt{x^2 - y^2} \in S$.

We are required to determine the form of the set $S$ that satisfies both conditions.

### Step 1: Analyze the Set $S$

First, according to Condition 1, the element $1$ must be in the set $S$ and it is the smallest element of $S$. Thus, $S$ contains all real numbers greater than or equal to 1.

### Step 2: Explore Consequences of Condition 2

Consider $x, y \in S$ with $x > y$. Then:

$$\sqrt{x^2 - y^2} = \sqrt{(x-y)(x+y)}$$

For this expression to be a real number present in $S$, we need to ensure it evaluates to a real number greater than or equal to 1.

### Step 3: Construct the Set $S$

From condition 2, $\sqrt{x^2 - y^2}$ should remain in the set $S$ for all $x, y \in S$. Consider:
- If $x = 1$, then $y$ must equal 1 (since $x$ is the smallest and equal to 1 by Condition 1). Thus, $\sqrt{x^2 - y^2} = \sqrt{1 - 1} = 0$, which cannot be in $S$ as it’s less than 1.
- Hence, as any $x \in S$ is paired with the smallest $y = 1$, when $x > 1$, it follows that $\sqrt{x^2 - 1^2} = \sqrt{x^2 - 1}$ must be included in $S$.

### Conclusion on the Form of $S$

The set must therefore be consistent for all values larger than or equal to 1. Therefore, the set $S$ should contain all real numbers greater than or equal to 1:

$$S = [1, \infty).$$

It satisfies both conditions because any operation $\sqrt{x^2 - y^2}$ for $x, y \in S$ results in a number that also belongs to the interval $[1, \infty)$, and 1 is the smallest number in this interval.

Thus, the set $S$ is:

$$\boxed{[1, \infty)}$$