10. From $1,2,3, \cdots, 2003$, select $k$ numbers such that among the selected $k$ numbers, there are definitely three numbers that can form the side lengths of an acute triangle. Find the minimum value of $k$ that satisfies the above condition.
10. Since any 3 numbers chosen from $1,2,3,4,5,7,9,12,15,20,25,33,42,54,69,88,112,143,182,232,295,376$, $478,609,775,986,1255,1597$ cannot form an acute triangle, we have $k_{\min } \geqslant 29$. If there exist $1 \leqslant a_{1}<a_{2}<\cdots<a_{29} \leqslant 2003$, and $a_{i} \in \mathbf{N}^{*}$, such that any three of these numbers cannot form an acute triangle. Therefore, $a_{1} \geqslant 1, a_{2} \geqslant 2, a_{3} \geqslant 3, a_{4} \geqslant 4, a_{5} \geqslant 5, a_{6} \geqslant \sqrt{a_{4}^{2}+a_{5}^{2}}, a_{6} \geqslant 7$, similarly, $a_{7} \geqslant 9$, and so on, we get $a_{8} \geqslant 12$, $a_{9} \geqslant 15, a_{10} \geqslant 20, a_{11} \geqslant 25, a_{12} \geqslant 33, a_{13} \geqslant 42, \cdots, a_{24} \geqslant 609, a_{25} \geqslant 775, a_{26} \geqslant 986, a_{27} \geqslant 1255$, $a_{28} \geqslant 1597$, so $2003 \geqslant a_{29} \geqslant 2032$, which is a contradiction. The assumption is false, i.e., among any 29 numbers chosen from $1,2, \cdots, 2003$, there must exist 3 numbers that can form an acute triangle. Therefore, $k_{\min }=29$.