The answer is 20.

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Each segment is the base of a triangle $T_k = AP_kQ_k$ that is similar to $ABC.$ The area of $T_k$ equals $\frac{k}{2450}$ times the area of $ABC,$ so the length $P_kQ_k$ equals $\sqrt{\frac{k}{2450}}$ times $BC.$ This is $\sqrt{\frac{k}{2450}} \cdot 5 \sqrt{3} = \frac17 \sqrt{\frac{3k}{2}}.$ This is rational if and only if $3k/2$ is a square, which happens if and only if $k = 6n^2$ for some $n,$ where $1 \le k \le 2450.$ This inequality is satisfied for $n = 1, 2, \ldots, 20,$ so the answer is $\fbox{20}.$