The sum of the digits 1 to 9=45, which is divisible by 9. The 9's remainder (or digital root) has to be preserved during arithmetical operations. If, for example, we add 23 to 87 we get 110. The digital root for each operand is found by simply adding the digits of the number together: 2+3=5 and 8+7=15, we keep doing this until we have a single figure: 1+5=6, so the digital root of 87 is 6. The remainder after dividing by 9 equals the digit root. 23 divided by 9 is 2 rem 5; 87/9=9 rem 6. When numbers are added together the result has a digital root equal to the sum of the digital roots of the operands; so 5+6=11 and 1+1=2, so the result of 23+87=110 and 110 has a digital root of 2 (1+1+0=2).
Where is all this leading? Since we can only use the digits 1 to 9 once, we know that no matter what numbers we choose to make out of those digits the final result will have a digit root of 9 because 4+5=9. But the digital root of 100 is 1; therefore it is not possible to find numbers using the digits 1 to 9 once only such that the sum is 100, because the digital roots don't add up. The closest we can get is 99, which has a digital root of 9 because 9+9=18=1+8=9, matching the digital root of 45. One answer (not the only one) is: 59+12+3+4+6+7+8=99.