Answer: yes, one can.

Let $G$ be a gravicenter of the triangle $ABC$, and $A_1$, $B_1$, $C_1$ be the midpoints of the sides $BC$, $AC$, $AB$ respectively. Denote the sides and the medians of the triangle $ABC$ in the following way: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.

Suppose that one can paint three of six segments mentioned red and three others blue so that it is not possible to construct a triangle using the segments of the same color. Then the length of some red segment is not less than the sum of two other red segments; the same is true for blue segments.

So, there are two of six segments, the sum of which is not less than the sum of four others: $x + y \ge z + u + v + w$. We show, however, that this situation is not possible. Consider the following three cases.

1) Both $x$ and $y$ are some sides of $\triangle ABC$. Without loss of generality, let $a + b \ge c + d + e + f$. But from the triangles $BGC$ and $AGC$ we have $2f/3 + 2e/3 > a$ and $2f/3 + 2d/3 > b$. Summing all three inequalities we get $f/3 > c + d/3 + e/3$ which contradicts the triangle inequality for medians: $d + e > f$.

2) Both $x$ and $y$ are some medians of $\triangle ABC$. Without loss of generality, let $d + e \ge a + b + c + f$. But from the triangles $BGC_1$ and $AGC_1$ we have $f/3 + c/2 > 2e/3$ and $f/3 + c/2 > 2d/3$, or $f/2 + 3c/4 > e$ and $f/2 + 3c/4 > d$. It follows that $f + 3c/2 > d + e \ge a + b + c + f$, or $c/2 > a + b$, a contradiction.

3a) $x$ and $y$ are a side and a median of $\triangle ABC$ having the common endpoint. Without loss of generality, let $a+e \ge b+c+d+f$. But this inequality contradicts the inequalities $a < b+c$ and $e < d+f$.

3b) $x$ is a median of $\triangle ABC$ with the endpoint in the middle of the side $y$. Without loss of generality, let $a + d \ge b + c + e + f$. But this inequality contradicts the inequalities $a < b+c$ and $d < e+f$.