bump... 1 year and 0 solves

$n=2,3,4$. $n=1$: obviously not true. $n\ge 5$: Let $(a,b,c)=(1,2,3)$.Easy to know not true. $n=2,3,4$: WLOG, let $a=1,1\le b\le c<n$. If $c<b+1$: $(l,j,k)=(0,0,0)$ works. If $b+c>n$: $(l,j,k)=(1,0,0)$ works. If $n+c>nb$: $(l,j,k)=(1,1,0)$ works. If $c\ge b+1,b+c \le n,n+c \le nb$: $\frac{n+1}{2}c\ge \frac{n+1}{2}b+\frac{n+1}{2},nb\ge n+c\Rightarrow \frac{n-1}{2}(b+c)\ge \frac{3n+1}{2}$ $\Rightarrow \frac{n(n-1)}{2}\ge \frac{3n+1}{2}$, but we have $n=2,3,4$, contradiction. So our answer is $n=2,3,4$.

ostrowski wrote: $n=2,3,4$. $n=1$: obviously not true. $n\ge 5$: Let $(a,b,c)=(1,2,3)$.Easy to know not true. $n=2,3,4$: WLOG, let $a=1,1\le b\le c<n$. If $c<b+1$: $(l,j,k)=(0,0,0)$ works. If $b+c>n$: $(l,j,k)=(1,0,0)$ works. If $n+c>nb$: $(l,j,k)=(1,1,0)$ works. If $c\ge b+1,b+c \le n,n+c \le nb$: $\frac{n+1}{2}c\ge \frac{n+1}{2}b+\frac{n+1}{2},nb\ge n+c\Rightarrow \frac{n-1}{2}(b+c)\ge \frac{3n+1}{2}$ $\Rightarrow \frac{n(n-1)}{2}\ge \frac{3n+1}{2}$, but we have $n=2,3,4$, contradiction. So our answer is $n=2,3,4$. carteacao do krl kkkkkkkkkkkkkk