Find all functions $f : \mathbb{R} \to \mathbb{Z}$ which satisfy the conditions: 

$f(x+y) < f(x) + f(y)$

$f(f(x)) = \lfloor {x} \rfloor + 2$
1. **Analyzing the given conditions:**
   - We need to find all functions \( f : \mathbb{R} \to \mathbb{Z} \) that satisfy:
     1. \( f(x+y) < f(x) + f(y) \)
     2. \( f(f(x)) = \lfloor x \rfloor + 2 \)

2. **Exploring the second condition:**
   - From \( f(f(x)) = \lfloor x \rfloor + 2 \), we can infer that \( f \) maps integers to integers in a specific way.
   - Let \( f(0) = a \). Then \( f(f(0)) = f(a) = 2 \). This implies \( f(a) = 2 \).

3. **Considering \( f \) at integer points:**
   - Let \( f(1) = b \). We need to determine \( f \) at other integer points.
   - Assume \( f \) is linear for integers, i.e., \( f(n) = cn + d \) for some constants \( c \) and \( d \).

4. **Using the first condition:**
   - For \( x = y = 0 \), \( f(0+0) < f(0) + f(0) \) implies \( f(0) < 2f(0) \), so \( f(0) > 0 \).
   - For \( x = y = 1 \), \( f(1+1) < f(1) + f(1) \) implies \( f(2) < 2f(1) \).

5. **Exploring specific values:**
   - Let \( f(2) = 2a \). Then \( 2a < 2b \), so \( a < b \).
   - From \( f(f(0)) = 2 \), we have \( f(a) = 2 \). Since \( a \) is an integer, \( a \) must be odd.

6. **Determining \( a \) and \( b \):**
   - From \( f(0) = a \) and \( f(1) = b \), we have \( f(2) = 2a \) and \( f(2) < 2b \), so \( 2a < 2b \) implies \( a < b \).
   - Given \( f(f(0)) = 2 \), \( f(a) = 2 \), and \( a \) is odd, we need to find \( a \) and \( b \) such that \( a + b = 3 \).

7. **Solving for \( a \) and \( b \):**
   - Since \( a \) is odd and positive, the possible values for \( a \) are 1 and 3.
   - If \( a = 1 \), then \( b = 2 \) (since \( a + b = 3 \)).
   - If \( a = 3 \), then \( b = 0 \), but this contradicts \( f(0) > 0 \).

8. **Verifying the solution:**
   - If \( a = 1 \) and \( b = 2 \), then \( f(0) = 1 \) and \( f(1) = 2 \).
   - For \( f(x) = x + 1 \) for all \( x \in \mathbb{Z} \), we have \( f(f(x)) = f(x+1) = x + 2 \), which satisfies \( f(f(x)) = \lfloor x \rfloor + 2 \).

9. **Checking the first condition:**
   - For \( f(x) = x + 1 \), \( f(x+y) = x + y + 1 \) and \( f(x) + f(y) = (x + 1) + (y + 1) = x + y + 2 \).
   - Clearly, \( x + y + 1 < x + y + 2 \), so the first condition is satisfied.

10. **Conclusion:**
    - The function \( f(x) = x + 1 \) for all \( x \in \mathbb{Z} \) satisfies both conditions.
    - However, for \( x \in \mathbb{R} \), \( f(x) = \lfloor x \rfloor + 1 \) does not satisfy the first condition for non-integer values.

The final answer is \( \boxed{ f(x) = x + 1 } \) for all \( x \in \mathbb{Z} \).