So maybe we should look for a large non-base. I think the following is pretty large and is the largest for small numbers: 0,1,2...n together with -1,-2,..-m such that 1+2+3...+m<n Note that it has the desirable property that we can't add any more to it to make it a non base. Maybe an induction will work?

Does anyone have a solution, or at least know where I can find one? I am curious about this. So far I have only proven the induction step $P(n) \implies P(n+1)$ if $m(n+1)-m(n)=1$ (where $m(n)$ is the number in my previous post). The only other case is where $m(n+1)-m(n)=0$ but this seems difficult (if not impossible).

Let $r$ be the smallest integer such that $1+2+...+r\ge n$ and suppose $|P|= (n+1)+r$. This implies there are at least $r$ positive values in $P$, denote the set $P^+$ and its elements $p_1< p_2<...<p_m$ where $m\ge r$ Now consider the $r^{\text{th}}$ smallest term $p_r\ge r$. Let $A=\{-n,-n+1,...,0\}\cup \{p_r,p_r+1,...,n\}$. Since $|P\cap A| = n+2$, for any $1 \le m \le p_r$ there is a pair of elements in $P$ of the form $(-k+m,k)$ for some $k> p_r$. Moreover, since $k>p_r$ all pairs are disjoint from any elements in $P^+$ that are less than $p_r$ This means that we can form a set $S=\{p_1,p_2,...,p_{r-1}\}\cup \{-k+m,k\}$ where all elements in $S\subset P$ are distinct. Furthermore if we treat the pair $(-k+m,k)$ as simply the element $m$ that can take any value $\le p_r$ then $S$ can easily form all values in the range $[1,n]$ since $p_1+p_2+...+p_r \ge 1+2+...+r\ge n$. Therefore $P$ is a base subset on the interval $[1,n]$. By the same token, then negative values $P^-\subset P$ can form a base subset on $[-n,-1]$. With $|P|\ge n+1$ there are two elements with sum $0$. The above considerations imply that $P$ must be a base subset on $M$. Therefore, by Kamil9876's construction, the minimum value of $k$ is indeed $|P|=n+1+r$.

That's pretty good

Cheers. Finally I found a solution I am satisfied with; I edited my post above

any new solutions for this one?