Denoting by $A$ the complement of $P$ in $[1,100]$, you are asking how large can $A$ be given that it does not contain any $10$-term arithmetic progression. The more general question 

*How large can a set $A\subset[1,N]$ be given that it does not contain any $k$-term arithmetic progression?* 

is one of the central problems in combinatorial number theory. There is no chance to give a precise answer, as an "explicit" function of $N$ and $k$, and it quite likely that this is impossible already in your special situation, where $N=n^2$ and $k=n$.

Here is an argument which seems to show that if $P\subset[1,n^2]$ meets every $n$-term progression contained in $[1,n^2]$, then $|P|>n+n^c$ with a positive absolute constant $c$. It would be interesting to improve this estimate to something like $|P|>(1+c)n+O(1)$.

Write $K:=|P|$ and $P=\{p_1,\dotsc,p_K\}$ where $1\le p_1<\dotsb<p_K\le n^2$. We have $p_1\le n$ and $p_K\ge n^2-(n-1)$, whence $p_K-p_1\ge(n-1)^2$. Next, for any $d\in[1,n]$, the set $P$ contains an element form each residue class modulo $d$, and it follows that there are at most $K-d$ pairs of consecutive elements of $P$ with the difference equal to $d$; also, if $d>n$, then there are no such pairs at all. Let $\Delta:=K-n$, and let $a>0$ and $r$ be defined by
\begin{align*}
   K-1 &= \Delta+(\Delta+1)+\dotsb+(\Delta+(a-1))+r \\
       &= a\Delta+\frac{a(a-1)}2 + r,\quad 0\le r<\Delta+a. \tag{1}
\end{align*}
Since there are totally $K-1$ pairs of consecutive elements of $P$, of them at most $\Delta$ pairs at distance $n$, at most $\Delta+1$ pairs at distance $n-1$, etc, we conclude that
\begin{align*}
  p_K-p_1 &\le n\Delta+(n-1)(\Delta+1)+\dotsb+(n-(a-1))(\Delta+(a-1))+(n-a)r \\
          &= \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r.
\end{align*}
Recalling the estimate $p_K-p_1\ge(n-1)^2$, we get
  $$ \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r \ge (n-1)^2. \tag{2} $$
This leads to the following optimization problem: find the smallest $\Delta=\Delta(n)$ such that, with $K=n+\Delta$, the quantities $a$ and $r$ defined by (1) satisfy (2). The solution should certainly be possible, but messy. I have at least investigated this problem numerically, and the results look **quite amazing** to me (which even makes me suspecting that there can be a mistake in my calculations or programming). Specifically, here is the graph of $\log \Delta /\log n$:

[![log(Delta(n)/log(n)][1]][1]

I absolutely cannot see any reason for the graph to be that steep for $n$ small and then flat for $n$ large, but anyway, the quotients seem to converge to a limit $c\approx 0.44$ (I have in fact made the computations for $n$ up to $10^5$). If everything is correct, this means that $K=|P|>n+n^c$. 

[1]: https://i.sstatic.net/IPjuS.jpg