Let $\mathbb N$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f:\mathbb N\to \mathbb N$ satisfying:
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    [*] for any $x,a\in\mathbb N$, the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a=1$;
    [*] for all $x\in \mathbb N$, we have $|f(x)-cx|<2023$. 
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[i]Proposed by Sutanay Bhattacharya[/i]
We need to find all real numbers \( c \) for which there exists a function \( f: \mathbb{N} \to \mathbb{N} \) satisfying the following conditions:
1. For any \( x, a \in \mathbb{N} \), the quantity \( \frac{f(x+a) - f(x)}{a} \) is an integer if and only if \( a = 1 \).
2. For all \( x \in \mathbb{N} \), we have \( |f(x) - cx| < 2023 \).

Let's analyze these conditions step by step.

1. **Condition Analysis:**
   - The first condition implies that the difference quotient \( \frac{f(x+a) - f(x)}{a} \) is an integer only when \( a = 1 \). This suggests that the function \( f \) has a specific behavior when incremented by 1.
   - The second condition implies that \( f(x) \) is close to a linear function \( cx \) with a bounded deviation.

2. **Function Transformation:**
   - Define \( g(x) = f(x) - \lfloor c \rfloor x \). This transformation helps us focus on the fractional part of \( c \). Note that \( g(x) \) will also satisfy the given conditions because \( \lfloor c \rfloor x \) is a linear function and does not affect the integer difference quotient condition.

3. **Bounding \( g(x) \):**
   - Since \( |f(x) - cx| < 2023 \), we have \( |g(x) - \{c\}x| < 2023 \), where \( \{c\} \) denotes the fractional part of \( c \).

4. **Behavior of \( g(x) \):**
   - If \( \{c\} < \frac{1}{2} \), then \( g(x) \) is bounded within a range that grows slower than \( x \). This would imply that \( g(x) \) can only take a limited number of values as \( x \) increases, leading to a contradiction because \( g(x) \) must be injective to satisfy the first condition.
   - Similarly, if \( \{c\} > \frac{1}{2} \), then \( h(x) = x - g(x) \) would also be bounded within a range that grows slower than \( x \), leading to a similar contradiction.

5. **Special Case \( \{c\} = \frac{1}{2} \):**
   - Let \( c = n + \frac{1}{2} \) for some integer \( n \). Consider the function \( f(x) = \left\lfloor \frac{x}{2} \right\rfloor + nx + 1 \). We need to check if this function satisfies the given conditions.
   - For \( a = 1 \), \( \frac{f(x+1) - f(x)}{1} \) is an integer.
   - For \( a > 1 \), \( \frac{f(x+a) - f(x)}{a} \) is not an integer because the floor function introduces a non-linear behavior that disrupts the integer difference quotient.

6. **Verification:**
   - For \( f(x) = \left\lfloor \frac{x}{2} \right\rfloor + nx + 1 \), we have \( |f(x) - (n + \frac{1}{2})x| < 2023 \) because the floor function and the constant term ensure the deviation is bounded.

Thus, the only possible value for \( c \) is when the fractional part of \( c \) is \( \frac{1}{2} \).

\(\blacksquare\)

The final answer is \( \boxed{ c = n + \frac{1}{2} } \) for some integer \( n \).