Belarusian Mathematical Olympiad

Answer: $n=2$, $\{x_1,x_2\}=\{4,13\}$.

It is clear that there are no solutions for $n=1$. Hence consider the case $n\ge 2$. Let $s={\sum }_i{x}_i$, $S={\sum }_i{x}_i^2$. Then $100s-S=1515$ and hence $100s>1515>100s-s^2$. Taking into account that $s$ is a nonnegative integer, we obtain from these inequalities that $s\ge 16$ and $|50-s|\ge 32$, i.e. $s\ge 82$ or $16\le s\le 18$.

We shall now show that $s\ge 82$ cannot hold. Suppose it does hold, then among the numbers $x_i$ we can find two elements $50>x_k\ge x_l>1$ — if this were not true, then either there is some $x_i=50$, implying $$\sum _{i=1}^n{x}_i(100-x_i)\ge 50\cdot (100-50)>1515,$$ or in view of $s\ge 82$, we would have $x_i=1$ for at least $82-49=33$ indices $i$, implying $$\sum _{i=1}^n{x}_i(100-x_i)\ge 33\cdot 1\cdot (100-50)=33\cdot 50>1515.$$ Hence, we have $50>x_k\ge x_l>1$ for some $k\ne l$ and we shall prove that $x_k(100-x_k)+x_l(100-x_l)>(x_k+1)(100-(x_k+1))+(x_l-1)(100-(x_l-1))$.

Expanding, gathering similar terms and simplifying we get that this inequality is equivalent to the inequality $2x_k-2x_l>-2$, which is true since $x_k\ge x_l$. So, replacing $x_k$ and $x_l$ by $x_k+1$ and $x_l-1$, we decrease the sum ${\sum }_i{x}_i(100-x_i)$ and after a finite number of such steps either one of the elements $x_i$ becomes equal to 50 or we have $x_i=1$ for at least 33 indices $i$, yielding a contradiction in both cases as it is shown above.

Therefore, we have $16\le s\le 18$. Since $S=100s-1515$ is odd, we have $s$ is also odd because $s$ and $S$ have the same parity. So $s=17$ and $S=185$.

If $n=2$ we easily get the solution $\{x_1,x_2\}=\{4,13\}$.

Suppose now that $n\ge 3$. Then for any $k$ we have $x_k^2\le 185$ and $$185=x_k^2+\sum _{i\ne k}{x}_i^2<x_k^2+(17-x_k{)}^2$$ (since the sum of squares is less than the square of the sum), implying $x_k<14$ and $(x_k-4)(x_k-13)>0$. Therefore, all elements $x_i$ are not greater than 3. But then we have $S=\sum x_i^2\le n\cdot 9\le 17\cdot 9<185$, a contradiction.