The sum of the digits of a 4-digit number is subtracted from the number. The resulting number is always

Take the number 2345

Sum of digits:

$2 + 3 + 4 + 5 = 14$

Subtract from the original number:

$2345 - 14 = 2331$

$2331 \div 9 = 259$ with no remainder

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Take the number 1678

Sum of digits:

$1 + 6 + 7 + 8 = 22$

Subtract from the original number:

$1678 - 22 = 1656$

$1656 \div 9 = 184$ with no remainder

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Let the four digits be $a, b, c, d$ where $a$ is the thousands digit, $b$ is the hundreds digit, $c$ is the tens digit, and $d$ is the ones digit.

The 4-digit number can be written as:

$1000a + 100b + 10c + d$

Sum of digits:

$a + b + c + d$

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Number minus sum of digits:

$(1000a + 100b + 10c + d) - (a + b + c + d)$

$= 1000a - a + 100b - b + 10c - c + d - d$

$= 999a + 99b + 9c$

$= 9(111a + 11b + c)$

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The result is $9(111a + 11b + c)$, which is always a multiple of 9.

Therefore, the resulting number is always divisible by 9.