We will prove that by induction. It's obvious for $n=1,2,3$. Let it be true for $1,2,...,k$. We'll prove that it's also true for $k+1$. Assume that we can't do this. \[a_1+...,+a_{k+1}=3k\]If there exist some $a_i$ which is $3$, then we can do it by induction. There can't be $1$ and $2$ at the same time. If there exists $1$, there can't be $2,3$. There can't be three $1$. If there exists one $1$, then \[3k=a_1+...+a_k+1 \geq 4k+1\]Contradiction If there exist two $1$, then \[3k=a_1+...+a_{k-1}+2 \geq 4k-2\]\[2 \geq k\]Contradiction If there exists one $2$, then \[3k=a_1+...+a_k+2 \geq 4k+2\]Contradiction If there exist two $2$, then \[3k=a_1+...+a_{k-1}+4 \geq 4k\]Contradiction