Arrange arbitrarily $1,2,\ldots ,25$ on a circumference. We consider all $25$ sums obtained by adding $5$ consecutive numbers. If the number of distinct residues of those sums modulo $5$ is $d$  $(0\le d\le 5)$,find all possible values of $d$.
1. **Define the sequence and sums:**
   Let the numbers be \( a_i \) for \( 1 \leq i \leq 25 \). We extend the sequence cyclically by setting \( a_{i+25} = a_i \). Define the sums of 5 consecutive numbers as \( s_i = a_i + a_{i+1} + a_{i+2} + a_{i+3} + a_{i+4} \).

2. **Case \( d = 1 \):**
   If \( d = 1 \), all sums \( s_i \) must be congruent to the same residue modulo 5. Consider the sequence \( a_i = i \). Then:
   \[
   s_i = a_i + a_{i+1} + a_{i+2} + a_{i+3} + a_{i+4} = i + (i+1) + (i+2) + (i+3) + (i+4) = 5i + 10
   \]
   Since \( 5i + 10 \equiv 0 \pmod{5} \), all \( s_i \equiv 0 \pmod{5} \). Thus, \( d = 1 \) is possible.

3. **Case \( d = 3 \):**
   Modify the previous sequence by swapping \( a_5 \) and \( a_6 \) such that \( a_5 = 6 \) and \( a_6 = 5 \). This changes \( s_1 \) and \( s_6 \):
   \[
   s_1 = 1 + 2 + 3 + 4 + 6 = 16 \equiv 1 \pmod{5}
   \]
   \[
   s_6 = 5 + 7 + 8 + 9 + 10 = 39 \equiv 4 \pmod{5}
   \]
   The other sums remain \( 0 \pmod{5} \). Thus, the residues are \( 0, 1, 4 \), and \( d = 3 \) is possible.

4. **Case \( d = 4 \):**
   Further modify the sequence by swapping \( a_{10} \) and \( a_{12} \) such that \( a_{10} = 12 \) and \( a_{12} = 10 \). This changes \( s_6, s_7, s_{11}, \) and \( s_{12} \):
   \[
   s_6 = 5 + 7 + 8 + 9 + 12 = 41 \equiv 1 \pmod{5}
   \]
   \[
   s_7 = 6 + 8 + 9 + 10 + 11 = 44 \equiv 4 \pmod{5}
   \]
   \[
   s_{11} = 10 + 11 + 12 + 13 + 14 = 60 \equiv 0 \pmod{5}
   \]
   \[
   s_{12} = 11 + 12 + 13 + 14 + 15 = 65 \equiv 0 \pmod{5}
   \]
   The residues are \( 0, 1, 2, 3 \), and \( d = 4 \) is possible.

5. **Case \( d = 5 \):**
   Arrange the sequence such that \( a_i \equiv 0 \pmod{5} \) for \( 1 \leq i \leq 4 \) and \( a_i \equiv 1 \pmod{5} \) for \( 5 \leq i \leq 9 \). Then:
   \[
   s_i \equiv i \pmod{5} \text{ for } 1 \leq i \leq 5
   \]
   Thus, \( d = 5 \) is possible.

6. **Case \( d = 2 \):**
   Suppose \( d = 2 \). Let the residues be \( m \) and \( n \) modulo 5. Assume \( s_1 \equiv s_2 \equiv m \pmod{5} \) and \( s_3 \equiv n \pmod{5} \). Consider the sum of all \( s_i \):
   \[
   \sum_{i=1}^{25} s_i = \sum_{i=1}^{25} a_i = \frac{25 \cdot 26}{2} = 325 \equiv 0 \pmod{5}
   \]
   Since the sum of five terms each congruent to \( m \) or \( n \) modulo 5 must be a multiple of 5, all terms must have the same residue modulo 5. This contradicts the assumption that there are two distinct residues. Hence, \( d = 2 \) is not possible.

The final answer is \( \boxed{ d = 1, 3, 4, 5 } \).