For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Given the 4-digit number:

Let the number be $abcd$.

Given:

$a+b+c=14$

$b+c+d=15$

$c=d+4$

To find the maximum number that satisfies the conditions, we need to maximize $a$, the thousands digit.

Substituting $c=d+4$ in $b+c+d=15$:

$b+(d+4)+d=15$

$b+2d+4=15$

$b+2d=11$

Subtracting $a+b+c=14$ from $b+c+d=15$:

$(b+c+d)-(a+b+c)=15-14$

$d-a=1$

$d=a+1$

Since $c$ must be a single digit less than or equal to 9, let $c=9$. Thus, $d=5$ and $d+4=9$.

Using $d=5$, we get:

$a=4$

$b+2d=11$

$b+2\times 5=11$

$b=1$

Thus, the digits of the number are:

$a=4$, $b=1$, $c=9$, $d=5$

The highest four-digit number satisfying the conditions is 4195.