Kezer wrote: A sum of $335$ pairwise distinct positive integers equals $100000$. a) What is the least number of uneven integers in that sum? b) What is the greatest number of uneven integers in that sum? Consider $n$ odd positive integers and $335-n$ even positive integers. In order the sum be even, we need $n$ even. We also have $100000\ge \sum_{k=1}^n(2k-1)+\sum_{k=1}^{335-n}2k$ $=n^2+(335-n)(336-n)$ So $2n^2-671n+12560\le 0$ and $20\le n\le 314$ (remembering that $n$ must be even And these two bounds indeed may be reached : $n=314$ with for example $\{1,3,5,7,...,627\}$ and $\{2,4,6,...,40,984\}$ $n=20$ with for example $\{1,3,5,7,..., 39\}$ and $\{2,4,6,...,626,628,690\}$ Hence the answer : $\boxed{n\ge 20\text{ and }n\le 314}$