a) Let $n = a q_1 + r_1 = b q_2 + r_2$, $a, b \in \mathbb{N}$, $a < b$, $r_1 \neq 0$, $r_2 \neq 0$. Suppose that $n = r_1 + r_2$. Then $n = a q_1 + r_1 = b q_2 + r_2 = r_1 + r_2$, ($q_1 \ge 0$, $q_2 \ge 0$) and $r_1 < a$, $r_2 < b$, so $b q_2 = r_1 < a \le b$. The inequality $b q_2 < b$ holds only if $q_2 = 0$. But then $r_1 = b q_2 = b \cdot 0 = 0$, a contradiction.

b) Let, by condition,
$$n = 29 q_1 + r_1 = 39 q_2 + r_2 = 59 q_3 + r_3 = r_1 + r_2 + r_3,$$
$r_1 < 29$, $r_2 < 39$, $r_3 < 59$. From these inequalities it follows that $59 q_3 = r_1 + r_2 \le 28 + 38 = 66$, so $q_3 = 1$. Then
$$r_1 + r_2 = 59. \quad (1)$$
Further, $39 q_2 = r_1 + r_3 \le 28 + 58 = 86$, so $q_2 \le 2$.
Consider two cases:
1) $q_2 = 1$. Then
$$r_1 + r_3 = 39. \quad (2)$$
$98 = 59 + 39 = r_1 + r_2 + r_1 + r_3 = n + r_1 = 29 q_1 + 2 r_1,$
i.e., $29 q_1 + 2 r_1 = 98$, so $q_1$ is even and $29 q_1 < 98$, hence $q_1 = 2$. Then $r_1 = \frac{1}{2}(98 - 2 \cdot 29) = 20$. But from (1) it follows that $r_2 = 59 - 20 = 39$, which is impossible since $r_2 < 39$.

2) $q_2 = 2$. Then
$$r_1 + r_3 = n - r_2 = 39 q_2 = 78. \qquad (3)$$
Then from (1) and (3) we obtain $29 q_1 + 2 r_1 = 137$, so $q_1$ is odd and $29 q_1 < 137$, hence either $q_1 = 1$ or $q_1 = 3$. If $q_1 = 1$, then $2 r_1 = 108$, i.e., $r_1 = 54$, a contradiction. If $q_1 = 3$, then $r_1 = 25$, $r_2 = 34$, $r_3 = 53$, and $n = 25 + 34 + 53 = 112$, which satisfies the problem condition.