Let, by condition,
$$n = 29q_1 + r_1 = 41q_2 + r_2 = 59q_3 + r_3 = r_1 + r_2 + r_3,$$
$r_1 < 29$, $r_2 < 41$, $r_3 < 59$. From these inequalities it follows that $59q_3 = r_1 + r_2 \le 28 + 40 = 68$, so $q_3 = 1$. Then
$$r_1 + r_2 = 59. \quad (1)$$

Further, $41q_2 = r_1 + r_3 \le 28 + 58 = 86$, so $q_2 \le 2$. Consider two cases:
$$1) \quad q_2 = 1. \quad \text{Then} \qquad r_1 + r_3 = 41. \qquad (2)$$
From (1) and (2) it follows that
$$100 = 59 + 41 = r_1 + r_2 + r_1 + r_3 = n + r_1 = 29q_1 + 2r_1,$$
i.e., $29q_1 + 2r_1 = 100$. Hence $q_1$ is even and $29q_1 < 100$, so $q_1 = 2$. Then $r_1 = \frac{1}{2}(100 - 2 \cdot 29) = 21$, $r_2 = 38$, $r_3 = 20$, and $n = 21 + 38 + 20 = 79$, which satisfies the problem condition.

2) $q_2 = 2$. Then
$$r_1 + r_3 = n - r_2 = 41q_2 = 82. \qquad (3)$$
From (1) and (3) it follows that $29q_1 + 2r_1 = 141$, so $q_1$ is odd and $29q_1 < 141$. Hence either $q_1 = 1$ or $q_1 = 3$. If $q_1 = 1$, then $2r_1 = 122$, i.e., $r_1 = 61$, a contradiction. If $q_1 = 3$, then $r_1 = 27$, $r_2 = 32$, $r_3 = 55$, and $n = 27 + 32 + 55 = 114$, which satisfies the problem condition.