$n=1112$ take $t_{0}=0.1111$ interesting numbers that are smaller that $t_{0}$ begin with four zeroes so they are smaller than $0.0001$ let $t_{0}=\sum_{i=1}^{n}a_{i}$ where $a_{i}$ are interesting numbers.now we have $t_{0}=\sum_{i=1}^{n}a_{i}<\sum_{i=1}^{n}0.0001=n*0.0001$ so $n>1111$ or $n\geq 1112$ take any $t$ such that $0<t\leq t_{0}$. take $b=\frac{t}{1112}$ we know that $b<0.0001$ and that $t=\sum_{i=1}^{1112}b$ so all we have to do is turn the $b$'s into different irrationals. take $\phi$ such that $b-556\phi>0$ and $b+556\phi<0.0001$. if $b$ is rational pick $\phi$ irrational and vice versa. the numbers $b-556\phi,b-555\phi,...,b-\phi,b+\phi,...,b+556\phi$ satisfy. if $t>t_{0}$ write it in this form $t=k*t_{0}+t^{'}$ where $k$ is an integer and $0<t^{'}<t_{0}$ and take one number from the representation of $t^{'}$ and add $k*t_{0}$ to it.