Pigeonholes?

Let's arrange the $19$ distinct numbers in ascending order, i.e. $\{x_{1}, x_{2}, ..., x_{19}\}$ where $x_{1} \lt x_{2} ... \lt x_{19}$ .

We need to prove that if we pick the top $9$ numbers and if their sum is $\lt 900$ , then the sum of all the $19$ numbers (irrespective of the value of lower $10$ numbers) will always be $\lt 1800$ .

In other words, for the sum of these $19$ numbers to be $\geq 1800$ , the sum of the top $9$ should be $\geq 900$

Let's choose the top $9$ elements, i.e. $\{x_{11}, x_{12}, ..., x_{19}\}$

One of the many sets of $9$ distinct integers that has maximum sum $\lt 900$ is $\{95, 96, 97, 98, 99, 100, 101, 102, 111\}$ (sum $= 899$ ).

Note that there are many other sets such that the $x_{11}$ number is $95$ and yet the sum is still $899$ . However, $x_{11}$ can't be $\gt 95$ for the sum to be $\lt 900$ . For the purpose of this proof, we don't need to treat those other sets differently than above set. We are mainly interested in the $x_{11}$ number.

Any other set of $9$ distinct integers is going to have only lesser sum and will yield overall (for all $19$ numbers) lesser sum than these $9$ numbers. Hence we choose these $9$ numbers to proceed with our proof.

The lower $10$ numbers can only be lesser than $95$ . The lower $10$ numbers with maximum sum that we can choose are $\{85, 86, 87, 88, 89, 90, 91, 92, 93, 94\}$ . Their sum $= 895$ .

Hence, if the sum of top $9$ numbers of our set of $19$ numbers is $\lt 900$ , then the sum of all $19$ numbers will always be $\lt 1800$ . If the sum of top $9$ numbers of our set is $\geq 900$ , then we anyways have the solution.

We still need to prove that $X$ can't be less than $9$ . For the given set below, the sum of all $19$ numbers is 1805 ( $\geq 1800$ ) and yet the sum of top $8$ numbers is 804 ( $\lt 900$ ).

$\{86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104\}$ .

Hence, X can't be less than $9$ for the given problem statement to be always true and sum of top $9$ numbers has to be $\geq 900$ for the sum of $19$ numbers to be $\geq 1800$ .

Key here is identifing critical case ; numbers are consecutive

 First number be 'n'
 19(n+9)>=1800 (1805 as 'n' is integer)

n=86

Now check whether 8 numbered subset is possible

max possible sum of 8 numbers is 804; so not possible

9 numbered set is possible as 96(next least number)+804=900

So 9 is minimal value for X.