Using 1-9 digits

Designate the integers sought by N, 2N, 3N. Since an integer yields the same remainder upon division by 9 as does the sum of its digits, the sum N + 2N + 3N must yield the same remainder upon division by 9 as does: 1 + 2 + 3 + … + 9 = 45 in order to meet the condition imposed by the problem. Hence, 6N (and consequently 3N) is divisible by 9.

Since 3N is to be a three-digit number, the first digit of N cannot exceed 3. It follows that the last digit of N cannot be 1, since the integer 2N would end with 2 and 3N would end with 3, and then none of these three digits is available to begin N. The integer N cannot terminate with 5, since 2N would end with 0. Also note that the first digit of 3N cannot be 1 or 2 otherwise N will be 2 digits.

a. If final digit of N is 2:

The final digits of 2N and 3N are, respectively, 4 and 6. The remaining two digits for 3N can be chosen only from 1, 3, 5, 7, 8 and 9. Since the sum of all the digits of 3N must be a multiple of 9, it follows that the first two digits of 3N are either 3 and 9 or 5 and 7. The possibilities for 3N are: 396, 576, 756, 936.

If 3N = 396  N will be: 132 to be rejected because the digit 3 is repeated.

If 3N = 576  N will be: 192 and 2N will be 384. This is OK.

If 3N = 756  N will be: 252 to be rejected because the digit 5 is repeated.

b. If final digit of N is 3:

The final digits of 2N and 3N are, respectively, 6 and 9. The remaining two digits for 3N can be chosen only from 1, 2, 4, 5, 7, and 8. It follows that the first two digits of 3N are either 1 and 8, 2 and 7 or 4 and 5. The possibilities for 3N are: 459, 549, 639, 729 or 819.

3N = 459  N will be: 153 to be rejected because the digit 5 is repeated.

3N = 549  N will be: 183 and 2N will be 366 to be rejected because the digit 3 is repeated.

3N = 639  N will be: 213 to be rejected because the digit 3 is repeated.

3N = 729  N will be: 243 to be rejected because the digit 2 is repeated.

3N = 819  N will be: 273 and 2N will be 546. This is OK.

c. If final digit of N is 4:

The final digits of 2N and 3N are, respectively, 8 and 2. The remaining two digits for 3N can be chosen only from 1, 3, 5, 6, 7, and 9. It follows that the first two digits of 3N are either 1 and 6 or 7 and 9. The possibilities for 3N are: 612, 792 or 972.

3N = 612  N will be: 204 to be rejected because the digit 2 is repeated.

3N = 792  N will be: 264 to be rejected because the digit 2 is repeated.

3N = 972  N will be: 324 to be rejected because the digit 2 is repeated.

d. If final digit of N is 6: The final digits of 2N and 3N are, respectively, 2 and 8. The remaining two digits for 3N can be chosen only from 1, 3, 4, 5, 7, and 9. It follows that the first two digits of 3N are either 1 and 9 or 3 and 7. The possibilities for 3N are: 378, 738 or 918.

3N = 378  N will be: 126 and 2N will be 252 to be rejected because the digit 2 is repeated.

3N = 738  N will be: 246 and 2N will be 492 to be rejected because the digit 2 is repeated.

3N = 918  N will be: 306 to be rejected because of the digit 0.

e. If final digit of N is 7: The final digits of 2N and 3N are, respectively, 4 and 1. The remaining two digits for 3N can be chosen only from 2, 3, 5, 6, 8, and 9. It follows that the first two digits of 3N are either 2 and 6, 3 and 5 or 8 and 9. The possibilities for 3N are: 351, 531, 621, 891 or 981.

3N = 351  N will be: 117 to be rejected because the digit 1 is repeated.

3N = 531  N will be: 177 to be rejected because the digit 7 is repeated.

3N = 621  N will be: 207 to be rejected because of the digit 0.

3N = 891  N will be: 297 to be rejected because the digit 9 is repeated.

3N = 981  N will be: 327 and 2N will be 654. This is OK.

f. If final digit of N is 8: The final digits of 2N and 3N are, respectively, 6 and 4. The remaining two digits for 3N can be chosen only from 1, 2, 3, 5, 7, and 9. It follows that the first two digits of 3N are either 2 and 3 or 5 and 9. The possibilities for 3N are: 324, 594 or 954.

3N = 324  N will be: 108 to be rejected because of the digit 0.

3N = 594  N will be: 198 to be rejected because the digit 9 is repeated.

3N = 954  N will be: 318 and 2N will be 636 to be rejected because the digit 3 is repeated.

g. If final digit of N is 9: The final digits of 2N and 3N are, respectively, 8 and 7. The remaining two digits for 3N can be chosen only from 1, 2, 3, 4, 5, and 6. It follows that the first two digits of 3N are only 5 and 6. The possibilities for 3N are: 567 or 657.

3N = 567  N will be: 189 and 2N will be 378 to be rejected because the digit 7 is repeated.

3N = 657  N will be: 219 and 2N will be 438. This is OK.

From all of above we deduce that the 4 possibilities for are N, 2N and 3N are: 192, 384, 576 or 219, 438, 657 or 273, 546, 819 or 327, 654, 981.

This will give a total of: 1152, 1314, 1638 and 1962 respectively.

Since Mod (1152, 162) = Mod (1314, 162) = Mod (1638, 162) = Mod (1962, 162) = 18

$Answer = \boxed{18}$

Let the numbers are $x,2x,3x$ .Sum is $6x$ .Define this numbers as :

$x=\overline{abc} \\ 2x=\overline{lmn} \\ 3x=\overline{pqr}$ .

After adding

$6x=\overline{abc}+ \overline{abc} + \overline{pqr} \\ \implies 6x=100a+10b+c+100l+10m+n+100p+10q+r \\ \implies 6x=99a+99l+99p+9b+9m+9q+a+b+l+m+p+q+c+n+r \\ \implies 6x=99(a+l+p) +9(b+m+q) + (1+2+3+4+5+6+7+8+9) \\ \implies 6x=99(a+l+p) + 9(b+m+q) +45 \\ \implies 2x=3[11(a+l+p) + (l+m+q)+5]$

.It clearly states that $3|x$ i.e. $3 | (a+b+c)$

It is clear $a=1,2,3$ and $x$ can not have $1,5$ as unit's digit.

Case(1) $a=1$ so,remembering $3 | (a+b+c)$ $(b,c)$ can be $(2,3);(1,4);(1,7);.....$

Case(2): $a=2$ so,remembering $3 | (a+b+c)$ $(b,c)$ can be $(1,3);(1,6);(2,5);.....$

Case(3): $a=3$ so,remembering $3 | (a+b+c)$ $(b,c)$ can be $(2,1);(1,5);(2,4);.....$

From checking these three cases the solutions will be same as stated by @Karim Fawaz

Note There is a something important that in each case when sum of three numbers is divided by $162$ remainder is $18$ .Why this is happening ????