There can be $0$, $1$ or $2$ balls of mass $0.38$ in a box since $3 \cdot 0.38 > 1$. In these three cases the box can contain at most $\lfloor \frac{1}{0.038} \rfloor = 26$, $\lfloor \frac{1-0.38}{0.038} \rfloor = 16$ and $\lfloor \frac{1-2 \cdot 0.38}{0.038} \rfloor = 6$ balls with mass $0.038$ respectively. If looking for the minimum number of boxes, we may therefore assume that there are only boxes of three kinds: with $0$ heavy and $26$ light balls; with $1$ heavy and $16$ light balls; with $2$ heavy and $6$ light balls. Let there be $x_0$, $x_1$ and $x_2$ boxes of each kind respectively. In order that all balls be packed, it is necessary and sufficient that $26x_0 + 16x_1 + 6x_2 \ge 5000$ and $x_1 + 2x_2 \ge 1000$. Multiply the second inequality by $10$ and add it to the first one. This gives $26(x_0 + x_1 + x_2) \ge 15000$, hence $x_0 + x_1 + x_2 \ge \frac{15000}{26} = 576.9...$. Because $x_0 + x_1 + x_2$ is an integer, it follows that $x_0 + x_1 + x_2 \ge 577$. So $577$ boxes are necessary.

Now let $x_0 = 0$, $x_1 = 154$, $x_2 = 423$. Then $x_0 + x_1 + x_2 = 577$, $x_1 + 2x_2 = 1000$ and $26x_0 + 16x_1 + 6x_2 = 5002 > 5000$. These relations show that $577$ boxes are sufficient. The answer is $577$.