I do not think this is possible.Proof
0 + 1 + 2 + 3 + ... + 8 + 9 = 45
In the sum we are looking for there are only 1- and 2-digit numbers.
Assume there is a decimal number ab, i.e. 10a + b.
The presence of ab modifies the sum above, i.e., 45, by
10a + b - a - b = 9a
which is a factor of 9. But 45 itself is divisible by 9, implying that any sum in which all 10 digits appear joined in at most a 2-digit number is always divisible by 9.
Actually this is clear from the fact that "the sum of digital roots is the digital root of the sum."