Find the number of all $\text{permutations}$ of $\left \{ 1,2,\cdots ,n \right \}$ like $p$ such that there exists a unique $i \in \left \{ 1,2,\cdots ,n \right \}$ that :

$$p(p(i)) \geq i$$
1. We need to find the number of permutations \( p \) of the set \(\{1, 2, \ldots, n\}\) such that there exists a unique \( i \in \{1, 2, \ldots, n\} \) for which \( p(p(i)) \geq i \).

2. Let's analyze the condition \( p(p(i)) \geq i \). For simplicity, let's start with \( i = 1 \):
   - If \( p(p(1)) \geq 1 \), then \( p(p(1)) = 1 \) or \( p(p(1)) > 1 \).
   - If \( p(p(1)) = 1 \), then \( p \) must map some element to 1, and that element must map back to 1. This implies \( p(1) = 1 \), which is a fixed point.

3. Now consider \( i = 2 \):
   - If \( p(p(2)) < 2 \), then \( p(p(2)) = 1 \) because \( p(p(2)) \) cannot be 2 (as it would violate the uniqueness condition).

4. Generalizing this, for \( i \geq 2 \):
   - \( p(p(i)) < i \) implies \( p(p(i)) = i-1 \).

5. Let's assume \( p(p(1)) = n \):
   - This means \( p(x) = n \) for some \( x \).
   - Since \( p(p(x)) = x-1 \), we have \( p(n) = x-1 \).
   - Also, \( p(1) = x \).

6. From \( p(p(n)) = n-1 \), we deduce \( p(x-1) = n-1 \).

7. By induction, we can show that \( p(x-i) = n-i \) for all \( i \leq x-1 \).

8. For \( p(n-(i+1)) = x-2 \), we need to ensure that for every chosen \( x \), \( p(j) \) is unique for every other number \( j \).

9. From the identities proven, \( p(x-(x-1)) = n-(x-1) \) implies \( p(1) = n-x+1 \).

10. Since \( p(1) = x \), we have \( x = n-x+1 \), leading to \( x = \frac{n+1}{2} \).

11. Therefore, for odd \( n \), \( x = \frac{n+1}{2} \) is an integer, and there is exactly one solution.

12. For even \( n \), \( x = \frac{n+1}{2} \) is not an integer, and there are no solutions.

The final answer is \( \boxed{ 1 } \) for odd \( n \) and \(0\) for even \( n \).