As we have been provided with 5 integers, they are $1,2,3,4,5$ among which 4 integers are included and 1 is excluded to get possible sum of $6,7,8,9$ from any 3 integers.

Step1 : Smallest possible sum of $3$ integers is $6$ which can obviously be obtained from $1,2,3$ as $4,5$ makes a difference of $6-4= 2 = (1+1) \phantom{cccccc} 6-5 = 1+0$ respectively. So we can easily decide that among $4$ integers $3$ of them are $1,2,3$

Step 2 : Second smallest possible sum is $7$ which be obtained by adding $1,2,4$ as 4 makes a difference of $7-4=3 = 1+2$ however 5 makes a difference of $7-5= 2=1+1$ .

Step3 : Next possible sums are $8$ and $9$ . Since $8=1+3+4$ and $9=5+4 =(2+3)+4 = 2+3+4$ (sum of three integers)

Hence we can see that $1,2,3,4$ are required integers.

Let the four numbers be $a, b, c, d$ ,

Hence the sum of any 3 of the numbers are different, we can know that the four numbers are different. ( If $a$ equal to $b$ , $a+c+d$ will equal to $b+c+d$ )

Let us say that $a+b+c=6\\a+b+d=7\\a+c+d=8\\b+c+d=9$ ,

add them up, we get $3a+3b+3c+3d=6+7+8+9$

$3(a+b+c+d)=30$

$a+b+c+d=10$

$1+2+3+4=10$ $\Rightarrow$ the four numbers cannot be $\boxed 5$

Let, the $4$ numbers are $a,b,c,d$

Now we can say, $a+b+c=6\\a+b+d=7\\a+c+d=8\\b+c+d=9$

Solving that we get, $a=1; b=2; c=3; d=4$

So $\boxed5$ is the answer, which is not to you.