OpenAI ChatGPT wrote: We will show that no such 100 positive distinct integers exist. Suppose that such 100 positive distinct integers $a_1, a_2, \ldots, a_{100}$ exist. Let $a_i^3$ be the cube of the number $a_i$ for $1\leq i\leq 100.$ Then we have \[a_k^3=\sum_{i=1, i\neq k}^{100}a_i^3\]for some $k$ where $1\leq k\leq 100.$ Since $a_i$ are distinct positive integers, we can assume that $a_k$ is the largest among all the $a_i$. Then we have \[100a_k^3=\sum_{i=1}^{100}a_i^3.\] Now, let $S$ be the sum of the cubes of the first 100 positive integers, so $S=1^3+2^3+\cdots+100^3.$ It can be shown that \[S=\left(\frac{100(101)}{2}\right)^2=50^2\cdot101^2.\]Therefore, we have \[100a_k^3=\sum_{i=1}^{100}a_i^3\leq S,\]which implies that $a_k^3\leq \frac{S}{100}$. Thus, we have $a_k^3\leq 50^2\cdot101^2/100 < 260000.$ Since $a_k$ is an integer, this implies that $a_k\leq 60.$ However, since $a_k$ is the largest of the $a_i$, we have $a_k^3 \geq (a_1+ a_2+ \ldots +a_{99})^3$. Applying the Cauchy-Schwarz inequality to the sum $a_1+ a_2+ \ldots +a_{99}$, we get \[(a_1+ a_2+ \ldots +a_{99})^3 \leq (99)(a_1^3+a_2^3+ \ldots +a_{99}^3).\]Hence, we have \[a_k^3 \geq (99)(S-a_k^3) = 99S - 99a_k^3.\]Substituting $S=50^2\cdot101^2$ and simplifying, we obtain $a_k^3\geq 4{,}999{,}950{,}000.$ This contradicts the fact that $a_k\leq 60.$ Therefore, there are no 100 positive distinct integers satisfying the given condition.