Find how many positive integers $k\in\{1,2,\ldots,2022\}$ have the following property: if $2022$ real numbers are written on a circle so that the sum of any $k$ consecutive numbers is equal to $2022$ then all of the $2022$ numbers are equal.
To solve this problem, we need to determine the values of \( k \) such that if 2022 real numbers are written on a circle and the sum of any \( k \) consecutive numbers is equal to 2022, then all the 2022 numbers must be equal.

1. **Understanding the Problem:**
   Let the 2022 real numbers be \( a_1, a_2, \ldots, a_{2022} \). The condition given is that the sum of any \( k \) consecutive numbers is 2022. Mathematically, this can be written as:
   \[
   a_i + a_{i+1} + \cdots + a_{i+k-1} = 2022 \quad \text{for all } i.
   \]

2. **Deriving the Condition:**
   Consider the sums of \( k \) consecutive numbers starting from different positions:
   \[
   a_1 + a_2 + \cdots + a_k = 2022,
   \]
   \[
   a_2 + a_3 + \cdots + a_{k+1} = 2022.
   \]
   Subtracting these two equations, we get:
   \[
   a_1 - a_{k+1} = 0 \implies a_1 = a_{k+1}.
   \]
   Repeating this process, we find:
   \[
   a_2 = a_{k+2}, \quad a_3 = a_{k+3}, \quad \ldots, \quad a_{2022} = a_{k+2022}.
   \]

3. **Periodicity and GCD:**
   For the numbers to be equal, the indices must cycle through all positions. This implies that \( k \) must be such that \( k \) and 2022 are coprime (i.e., \(\gcd(k, 2022) = 1\)). This is because if \( \gcd(k, 2022) \neq 1 \), then the sequence will not cover all indices, and we will not be able to conclude that all numbers are equal.

4. **Counting the Valid \( k \):**
   We need to count the number of integers \( k \) in the range \( 1 \leq k \leq 2022 \) such that \(\gcd(k, 2022) = 1\). This is given by Euler's Totient Function \(\phi(2022)\).

5. **Calculating \(\phi(2022)\):**
   First, factorize 2022:
   \[
   2022 = 2 \times 3 \times 337.
   \]
   Using the formula for Euler's Totient Function:
   \[
   \phi(2022) = 2022 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 - \frac{1}{337}\right).
   \]
   Simplifying this:
   \[
   \phi(2022) = 2022 \times \frac{1}{2} \times \frac{2}{3} \times \frac{336}{337} = 1011 \times \frac{2}{3} \times \frac{336}{337}.
   \]
   \[
   \phi(2022) = 1011 \times \frac{672}{1011} = 672.
   \]

The final answer is \(\boxed{672}\).