A natural number is called [i]triangular[/i] if it may be presented in the form $\frac{n(n+1)}2$. Find all values of $a$ $(1\le a\le9)$ for which there exist a triangular number all digit of which are equal to $a$.
To solve the problem, we need to find all values of \( a \) (where \( 1 \le a \le 9 \)) for which there exists a triangular number whose digits are all equal to \( a \).

1. **Definition of Triangular Number:**
   A triangular number can be expressed as:
   \[
   T_n = \frac{n(n+1)}{2}
   \]
   where \( n \) is a natural number.

2. **Form of the Number:**
   We need to find if there exists a triangular number such that all its digits are equal to \( a \). This number can be written as:
   \[
   a \cdot \frac{10^k - 1}{9}
   \]
   where \( k \) is the number of digits.

3. **Equating the Two Forms:**
   We equate the triangular number to the form with all digits equal to \( a \):
   \[
   \frac{n(n+1)}{2} = a \cdot \frac{10^k - 1}{9}
   \]
   Multiplying both sides by 18 to clear the fractions:
   \[
   9n(n+1) = 2a(10^k - 1)
   \]

4. **Simplifying the Equation:**
   Rearrange the equation:
   \[
   9n(n+1) = 2a \cdot 10^k - 2a
   \]
   \[
   9n(n+1) + 2a = 2a \cdot 10^k
   \]

5. **Modulo 5 Analysis:**
   Consider the equation modulo 5:
   \[
   9n(n+1) + 2a \equiv 2a \cdot 10^k \pmod{5}
   \]
   Since \( 10 \equiv 0 \pmod{5} \), we have:
   \[
   9n(n+1) + 2a \equiv 0 \pmod{5}
   \]
   Simplifying further:
   \[
   4n(n+1) + 2a \equiv 0 \pmod{5}
   \]
   \[
   4n(n+1) \equiv -2a \pmod{5}
   \]
   \[
   4n(n+1) \equiv 3a \pmod{5}
   \]

6. **Possible Values of \( a \):**
   We need to check which values of \( a \) satisfy the above congruence. We test each \( a \) from 1 to 9:

   - For \( a = 1 \):
     \[
     4n(n+1) \equiv 3 \pmod{5}
     \]
     Testing small values of \( n \), we find \( n = 1 \) works:
     \[
     4 \cdot 1 \cdot 2 \equiv 3 \pmod{5}
     \]
     So, \( a = 1 \) is valid.

   - For \( a = 2 \):
     \[
     4n(n+1) \equiv 6 \equiv 1 \pmod{5}
     \]
     Testing small values of \( n \), we find no \( n \) works.

   - For \( a = 3 \):
     \[
     4n(n+1) \equiv 9 \equiv 4 \pmod{5}
     \]
     Testing small values of \( n \), we find \( n = 2 \) works:
     \[
     4 \cdot 2 \cdot 3 \equiv 4 \pmod{5}
     \]
     So, \( a = 3 \) is valid.

   - For \( a = 4 \):
     \[
     4n(n+1) \equiv 12 \equiv 2 \pmod{5}
     \]
     Testing small values of \( n \), we find no \( n \) works.

   - For \( a = 5 \):
     \[
     4n(n+1) \equiv 15 \equiv 0 \pmod{5}
     \]
     Testing small values of \( n \), we find \( n = 0 \) works:
     \[
     4 \cdot 0 \cdot 1 \equiv 0 \pmod{5}
     \]
     So, \( a = 5 \) is valid.

   - For \( a = 6 \):
     \[
     4n(n+1) \equiv 18 \equiv 3 \pmod{5}
     \]
     Testing small values of \( n \), we find \( n = 3 \) works:
     \[
     4 \cdot 3 \cdot 4 \equiv 3 \pmod{5}
     \]
     So, \( a = 6 \) is valid.

   - For \( a = 7 \):
     \[
     4n(n+1) \equiv 21 \equiv 1 \pmod{5}
     \]
     Testing small values of \( n \), we find no \( n \) works.

   - For \( a = 8 \):
     \[
     4n(n+1) \equiv 24 \equiv 4 \pmod{5}
     \]
     Testing small values of \( n \), we find \( n = 4 \) works:
     \[
     4 \cdot 4 \cdot 5 \equiv 4 \pmod{5}
     \]
     So, \( a = 8 \) is valid.

   - For \( a = 9 \):
     \[
     4n(n+1) \equiv 27 \equiv 2 \pmod{5}
     \]
     Testing small values of \( n \), we find no \( n \) works.

Conclusion:
The valid values of \( a \) are \( 1, 3, 5, 6, 8 \).

The final answer is \( \boxed{1, 3, 5, 6, 8} \).