Let us split $1000$ into $8$ numbers $100$ and one number $200$. In this case, the sum of some terms can have $9$ different values: $100 \cdot 1$, $100 \cdot 2$, $\ldots$, $100 \cdot 8$, and $200 + 100 \cdot 7 = 900$.

Let us prove that it is impossible to obtain less than $9$ different numbers. Since $9$ does not divide $1000$, a split of $1000$ results in at least $2$ different terms. Let us sort the terms in ascending order. For the list, we first pick only the $1$st term, then $1$st and $2$nd, then $1$st, $2$nd and $3$rd, and so on, until we pick the first $8$ numbers – in this way, we get $8$ different numbers. Now we take the last $8$ terms: this new sum is larger than any previous sum, therefore, there cannot be fewer than $9$ different numbers listed.