The chalkboard hadn't been cleaned. (Corrected once.)

The given condition on $x_n,x_p,x_m$ is that $x_n^2 < (x_p+x_m)^2$ and $(x_p-x_m)^2 < x_n^2$ , so that $x_n < x_p + x_m$ and $|x_p-x_m| < x_n$ , so that $x_n < x_p + x_m$ , $x_p < x_n + x_m$ and $x_m < x_p + x_n$ . Thus we want to find subsets of $\{1,2,3,...,1234\}$ such that no triplet of elements of the subset forms the side lengths of a proper triangle. Three numbers $a < b < c$ form the sides of a proper triangle precisely when $c < a + b$ .

Thus a collection of $\{x_1,x_2,...,x_q\}$ of numbers, with $x_1 < x_2 < ... < x_q$ , will be such that no triplet of its elements forms the side lengths of a proper triangle provided that $x_k \ge x_i + x_j$ whenever $i < j < k$ . This requires precisely that $x_k \ge x_{k-1} + x_{k-2}$ for all $3 \le k \le q$ .

We need to find how many positive integers $x_1 < x_2 < ... < x_q$ can be chosen so that $x_k \ge x_{k-1} + x_{k-2}$ for all $3 \le k \le q$ . Note that $x_1 \ge 1 = F_2$ and $x_2 \ge 2 = F_3$ , where $F_n$ is the $n$ th Fibonacci number (so $F_0=0$ , $F_1=1$ , $F_2=1$ , and so on). If it is true that $x_k \ge F_{k+1}$ for $1 \le k \le u$ , then $x_{u+1} \ge x_u + x_{u-1} \ge F_{u+1} + F_u = F_{u+2}$ . Thus we deduce that $x_u \ge F_{u+1}$ for all $1 \le u \le q$ , and so $F_{q+1} \; \le \; x_q \; \le \; 1234$ Since $F_{16} = 987$ and $F_{17} = 1597$ , we deduce that $q+1 \le 16$ , and hence $q \le 15$ .

Thus, if a subset of $\{1,2,3,...,1234\}$ of size $q$ contains no triplet that forms the sides of a proper triangle, then $q \le 15$ . We can find a set of size $15$ with this property, namely $\{F_n\,:\, 2 \le n \le 16\} \; = \; \{1,2,3,5,8,13,21,34,55,89,144,233,377,610,987\}$ Any set with $\boxed{16}$ or more elements must contain a triplet that forms the sides of a proper triangle.