 Solution

For each $k$, let ${\epsilon }_{n,k}=\left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)-\lfloor \left(\genfrac{}{}{0ex}{}{n}{k}\right)f(k/n)\rfloor$. The floor function ensures $0\le {\epsilon }_{n,k}<1$, even when the quantity being rounded is negative. The integer endpoint hypothesis gives ${\epsilon }_{n,0}={\epsilon }_{n,n}=0$. Thus the error of the rounded Bernstein polynomial satisfiesIt suffices to bound this unnormalized sum uniformly. If $0\le x\le 1/4$, put $q=x/(1-x)\le 1/3$ and sum a geometric series:The last inequality follows by maximizing $x(1-x{)}^n$, whose maximum occurs at $x=1/(n+1)$ and is at most $1/(n+1)$. Reflection gives the same bound for $3/4\le x\le 1$. In the middle interval both $x$ and $1-x$ are at most $3/4$, so each summand is at most $(3/4{)}^n$. Therefore, for $n\ge 2$,Endpoint integrality matters: at zero or one a noninteger value would leave a fixed nonzero rounding error.