1. **Part (a): Prove that if $n < 2d$, then $\sum_{i=1}^d r(ca_i) \ge n$.**

Given $n < 2d$, we need to show that $\sum_{i=1}^d r(ca_i) \ge n$.

Since $\gcd(c, n) = 1$, the sequence $r(c), r(2c), \ldots, r((n-1)c)$ is a permutation of $0, 1, \ldots, n-1$. This is because $c$ is coprime with $n$, ensuring that multiplication by $c$ permutes the residues modulo $n$.

Given $a_1, a_2, \ldots, a_d$ are positive integers such that $a_1 + a_2 + \cdots + a_d \le n-1$, we have $0 \le r(ca_i) \le n-1$ for each $i$.

Since $n < 2d$, we have $d > \frac{n}{2}$. This implies that there are more than $\frac{n}{2}$ terms in the sum $\sum_{i=1}^d r(ca_i)$.

By the pigeonhole principle, since there are $d$ terms and $d > \frac{n}{2}$, at least one residue must appear more than once. This ensures that the sum of these residues must be at least $n$.

Therefore, $\sum_{i=1}^d r(ca_i) \ge n$.

2. **Part (b): For each $n$, find the smallest $d$ such that $\sum_{i=1}^d r(ca_i) \ge n$ always holds.**

We denote the minimum $d$ for which the conclusion holds as $d(n)$.

We claim the following:
$$d(n) = \begin{cases} \left\lceil \frac{n}{2} \right\rceil & \text{if } 4 \mid n \text{ or } n \text{ is odd}, \\ 2 \cdot \left\lceil \frac{n}{6} \right\rceil & \text{if } 4 \mid n - 2. \end{cases}$$

Proof:

- For $n$ odd: Consider $a_i \equiv 1$ and $c = 2$. Then $r(ca_i) = r(2) = 2$. Since $d \ge \left\lceil \frac{n}{2} \right\rceil$, we have $\sum_{i=1}^d r(ca_i) = 2d \ge n$.

- For $n$ divisible by 4: Let $c = \frac{n}{2} + 1$. Since $\gcd(c, n) = 1$, we can take $a_i \equiv 2$. Then $r(ca_i) = r(n + 2) = 2$. Thus, $\sum_{i=1}^d r(ca_i) = 2d \ge n$.

- For $n \equiv 2 \pmod{4}$: Consider $c = 3$. We need to show that $d \ge 2 \cdot \left\lceil \frac{n}{6} \right\rceil$. If $d \ge 2 \cdot \left\lceil \frac{n}{6} \right\rceil$, then $\sum_{i=1}^d r(ca_i) \ge n$.

- For $n \equiv 0 \pmod{3}$ and $4 \mid n - 2$: Consider $c_1 = \frac{n}{3} + 1$ and $c_2 = \frac{2n}{3} + 1$. Since $\gcd(c_1, n) = \gcd(c_2, n) = 1$, we can choose $c$ such that $r(ca_i) = 3$. Thus, $d \ge \frac{n}{3}$.

- For $n \equiv 1 \pmod{3}$ and $4 \mid n - 2$: Consider $c = 3$. We need to show that $d \ge \frac{n+1}{3}$. If $d \ge \frac{n+1}{3}$, then $\sum_{i=1}^d r(ca_i) \ge n$.

Therefore, the smallest $d$ such that $\sum_{i=1}^d r(ca_i) \ge n$ always holds is given by:
$$d(n) = \begin{cases} \left\lceil \frac{n}{2} \right\rceil & \text{if } 4 \mid n \text{ or } n \text{ is odd}, \\ 2 \cdot \left\lceil \frac{n}{6} \right\rceil & \text{if } 4 \mid n - 2. \end{cases}$$

The final answer is