Given three non-negative real numbers $a$, $b$ and $c$. The sum of the modules of their pairwise differences is equal to $1$, i.e. $|a- b| + |b -c| + |c -a| = 1$. What can the sum $a + b + c$ be equal to?
1. Given the condition \( |a - b| + |b - c| + |c - a| = 1 \), we need to prove that \( a + b + c \geq \frac{1}{2} \).

2. Without loss of generality, assume \( a \leq b \leq c \). This assumption simplifies the absolute values:
   \[
   |a - b| = b - a, \quad |b - c| = c - b, \quad |c - a| = c - a
   \]
   Therefore, the given condition becomes:
   \[
   (b - a) + (c - b) + (c - a) = 1
   \]
   Simplifying this, we get:
   \[
   c - a + c - a = 1 \implies 2(c - a) = 1 \implies c - a = \frac{1}{2}
   \]

3. Since \( c - a = \frac{1}{2} \), we can express \( c \) in terms of \( a \):
   \[
   c = a + \frac{1}{2}
   \]

4. Now, consider the sum \( a + b + c \). Since \( a \leq b \leq c \), we have:
   \[
   a \leq b \leq a + \frac{1}{2}
   \]
   This implies \( b \) is between \( a \) and \( a + \frac{1}{2} \). Therefore, we can write:
   \[
   a \leq b \leq a + \frac{1}{2}
   \]

5. To find the minimum value of \( a + b + c \), consider the case when \( b \) is as small as possible, i.e., \( b = a \):
   \[
   a + b + c = a + a + (a + \frac{1}{2}) = 3a + \frac{1}{2}
   \]
   Since \( a \geq 0 \), the minimum value of \( 3a + \frac{1}{2} \) occurs when \( a = 0 \):
   \[
   3(0) + \frac{1}{2} = \frac{1}{2}
   \]

6. Therefore, the minimum value of \( a + b + c \) is \( \frac{1}{2} \). Since \( a, b, c \) are non-negative, \( a + b + c \) can be any value greater than or equal to \( \frac{1}{2} \).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1}{2} } \) or greater.