We have given ten positive integers $x_1, x_2, \ldots,x_{10}$ s.t. $(1) \;\; \sum_{k=1}^{10} x_k = 1000$, $(2) \;\; \prod_{i=1}^{10} x_k! = n^{10}$, where $n$ is a positive integer. We WLOG assume $x_1 \leq x_2 \leq \cdots \leq x_{10}$. Then $x_{10} \geq 100$ by equation (1). Hence, if $x_{10}=100$, then according to equation (1) we obtain $x_1 = x_2 = \cdots = x_{10} = 100$ and $n=100!$. Next assume $x_{10}>100$. Then $101 \mid x_{10}!$, yielding (since $101$ is a prime) $101 \mid n^{10}$ by equation (2), implying $101^{10} \mid n^{10} = \prod_{k=1}^{10} x_k!$. Therefore (since $x_k<1000$ by equation (1) give us $101^2 \nmid x_k$) implies $101 \mid x_k$ and $x_k \geq 101$ for all $k \in \{1,2,\ldots ,10\}$. Consequently $\sum_{k=1}^{10} x_k \geq 101 \cdot 10 = 1010$, contradicting equation (1). Conclusion: The only solution of the system of equations (1)-(2) in positive integers is $x_1 = x_2 = \cdots = x_{10} = 100$ and $n=100!$.

Let the positive integers be $a_1,a_2,...,a_{10}$. Claim: $a_i\leq 100 \quad \forall i$ Proof Let us assume that one of the $a_i$'s is $>100$. WLOG, let it be $a_1$. Then, $a_1\geq 101 \implies 101 \mid a_1!$. Given, $a_1!a_2!...a_{10}!=k^{10}$, $k \in \mathbb{Z}$ Then $101 \mid k^{10} \implies 101^{10}\mid k^{10}$, as $101$ is prime. If $101^2 \mid a_i$ for some $i$, then $a_i\geq 101^2 \implies a_i \geq 10201$, which contradicts the fact that the sum of all the $a_i$ is $1000$. Then, at most $101^1$ can divide any of the $a_i!$. As $101^{10} \mid k^{10}$, it must imply that for all $i$, $101 \mid a_i! \implies a_i \geq 101$ for all $i$. But then their sum $\geq 1010 <1000$, again which contradicts the problem statement. Therefore, our assumption was wrong and all the $a_i's\leq 100 \implies \;\; \sum_{i=1}^{10} a_i \leq 1000$, with equality only when $\boxed{a_1=a_2=\cdots=a_{10}=100}$, and we're done!

..............

Why $101\vert k^{10} \implies 101^{10} \vert k^{10}$? does this property only also holds with non-prime numbers?