Find all monic polynomials $P(x)=x^{2023}+a_{2022}x^{2022}+\ldots+a_1x+a_0$ with real coefficients such that $a_{2022}=0$, $P(1)=1$ and all roots of $P$ are real and less than $1$.
1. We are given a monic polynomial \( P(x) = x^{2023} + a_{2022}x^{2022} + \ldots + a_1x + a_0 \) with real coefficients such that \( a_{2022} = 0 \), \( P(1) = 1 \), and all roots of \( P \) are real and less than \( 1 \).

2. Since \( P(x) \) is monic and has real coefficients, we can write it as:
   \[
   P(x) = (x - r_1)(x - r_2) \cdots (x - r_{2023})
   \]
   where \( r_i \) are the roots of \( P(x) \).

3. Given that all roots are real and less than \( 1 \), we have \( r_i < 1 \) for all \( i \).

4. We are also given that \( a_{2022} = 0 \). By Vieta's formulas, the sum of the roots of \( P(x) \) is zero:
   \[
   r_1 + r_2 + \cdots + r_{2023} = 0
   \]

5. Additionally, we know that \( P(1) = 1 \). Substituting \( x = 1 \) into the polynomial, we get:
   \[
   P(1) = (1 - r_1)(1 - r_2) \cdots (1 - r_{2023}) = 1
   \]

6. Since \( r_i < 1 \) for all \( i \), each term \( (1 - r_i) \) is positive. Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   1 \leq \left( \frac{(1 - r_1) + (1 - r_2) + \cdots + (1 - r_{2023})}{2023} \right)^{2023}
   \]

7. Simplifying the expression inside the AM-GM inequality:
   \[
   1 \leq \left( 1 - \frac{r_1 + r_2 + \cdots + r_{2023}}{2023} \right)^{2023}
   \]

8. Since \( r_1 + r_2 + \cdots + r_{2023} = 0 \), we have:
   \[
   1 \leq 1
   \]

9. The equality condition in the AM-GM inequality holds if and only if all the terms are equal. Therefore, we must have:
   \[
   1 - r_1 = 1 - r_2 = \cdots = 1 - r_{2023}
   \]

10. This implies that \( r_1 = r_2 = \cdots = r_{2023} = 0 \).

11. Hence, the polynomial \( P(x) \) must be:
    \[
    P(x) = x^{2023}
    \]

The final answer is \( \boxed{ P(x) = x^{2023} } \).