Smallest set such that all arithmetic progression will always contain at least a number in a set 
Let $S= \left\{ 1,2,3,...,100 \right\}$ be a set of positive integers from $1$ to $100$. Let $P$ be a subset of $S$ such that any arithmetic progression of length 10 consisting of numbers in $S$ will contain at least a number in $P$. What is the smallest possible number of elements in $P$ ?  

Denote $|P|$ as the number of elements in $P$. We shall find the smallest possible value of $|P|$. 
For $|P|=16$, we have the answer by @RobertIsrael below.
However, for $|P|<16$, I can neither find such set $P$ nor prove that $|P|$ cannot be less than $16$. So my question is:

Is it true that $|P| \geq 16$? How can I prove it? If not, what is the minimum amount of elements in $P$ ?

Also, I am wondering that:

If we replace 10 with an even number $n$,and $100$ with $n^2$, can we find the minimum of $|P|$ ?

Any answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately.
 A: Considering the complement of $P$ in $[1,100]$, you are asking how large can a subset of $[1,100]$ be given that it does not contain any $10$-term arithmetic progression. The more general question 
How large can a subset of $[1,N]$ 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 showing that if $P\subset[1,n^2]$ meets every $n$-term progression contained in $[1,n^2]$, then $|P|>n+n^{0.5+o(1)}$. (See also the paragraph at the very end for the estimate $|P|\ge 12$ in your special case where $P\subset[1,100]$ and we want to block all $10$-term progressions.) It would be interesting to improve these estimates or at least to decide whether $|P|>Cn$ holds true with an absolute constant $C>1$.
Write $K:=|P|$, $\Delta:=K-n$, and $P=\{p_1,\dotsc,p_K\}$ where $1\le p_1<\dotsb<p_K\le n^2$. Notice that $p_1\le n$ and $p_K\ge n^2-(n-1)$, whence $p_K-p_1\ge(n-1)^2$. 
For any $d\in[1,n]$, the set $P$ contains an element from every 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 $a$ 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 $p_K-p_1\ge(n-1)^2$, and using ($1$), we get
\begin{align*}
  (n-1)^2 &\le \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r \\
    &= n\Big(a\Delta+\frac{a(a-1)}2 + r\Big) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar \\
    &= n(K-1) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar. \tag{2} 
\end{align*}
We now assume, aiming at a contradiction, that $\Delta<n^c$ with an absolute constant $0<c<0.5$. From (1) we get then
  $$ K-1 \ge \Delta a + \frac{a(a-1)}2 \ge \frac12\,a^2 - 1 $$
implying $a\le\sqrt{2K}$; hence, $\Delta a=O(n^{0.5+c})$ and $r=a+\Delta=O(n^{0.5})$. As a result,
  $$ \frac12\,a^2 = K-1+\frac12\,a-\Delta a - r > K - O(n^{0.5+c}), $$
leading to $a>(1-o(1))\sqrt{2K}$. 
With these estimates in mind, from (2) we obtain
  $$ n^2 + O(n) \le nK - \frac12\,\Delta a^2 - \frac13\,a^3;  $$
that is,
  $$ \Delta n \ge \frac12\,\Delta a^2 + \frac13\,a^3 + O(n). $$
Consequently, 
  $$ n^{1+c} \ge \Delta n \ge \frac13\,a^3 + O(n) \ge (1-o(1))(2K)^{1.5} + O(n) > n^{1.5} + O(n), $$
a contradiction.

As an illustration of this approach, let's show that one needs at least $12$ elements to block every $10$-term progression in $[1,100]$. Suppose for a contradiction that $P\subset[1,100]$ is an $11$-element set blocking all such progressions. There are $|P|-1=10$ pairs of consecutive elements of $P$. Of these ten pairs, there is at most one pair with distance $10$ between its two elements, at most two pairs with distance $9$, at most three pairs with distance $8$, and at most four pairs with distance $7$. Therefore the largest element of $P$ exceeds the smallest one by at most $1\cdot 10+2\cdot 9 + 3\cdot 8 + 4\cdot 7=80$. It follows that either the smallest element of $P$ is at least $11$, or its largest element is at most $90$; but then $P$ does not block at least one of the progressions $[1,10]$ and $[91,100]$, a contradiction.
A: Using a tabu search procedure, I have found a solution for $|P|=17$, namely ${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$.  I don't know if this is optimal. 
EDIT: Found a solution for $|P|=16$, namely 
$$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$
