Proof: For a finite multiset $S$ of integers, let $\sigma(S)$ denote the sum of all elements in $S$ (counting multiplicities), and let $\Sigma(S) = \{\sigma(T) \mid \sigma(T) \neq T \subseteq S\}$, where $\Sigma(S)$ is treated as a set (ignoring multiplicities). Let $\Sigma_n(S)$ denote the set of congruence classes modulo $n$ of the elements in $\Sigma(S)$. For $x \in \mathbb{Z}$, let $\bar{x}$ denote the congruence class of $x$ modulo $n$. Under this notation, conditions (2) and (3) can be written as
$$\Sigma_n(\{a_1, a_2, \dots, a_k\}) = \{\bar{1}, \bar{2}, \dots, \bar{\ell}\}.$$
Step 1: The main idea is to add an element $x$ to a nonempty multiset $S$, obtaining $T = S \cup \{x\}$. Assuming that neither $\Sigma_n(S)$ nor $\Sigma_n(T)$ contains $\bar{0}$, we compare $\Sigma_n(S)$ and $\Sigma_n(T)$. In particular, if $\Sigma_n(T)$ contains exactly one more element than $\Sigma_n(S)$, we can deduce a very refined structure for $\Sigma_n(S)$.
Observe that
$$\Sigma(T) = \Sigma(S) \cup (\Sigma(S) + x) \cup \{x\},$$
so $\Sigma_n(S) \subset \Sigma_n(T)$. Moreover, $\overline{\sigma(S) + x}$ belongs to $\Sigma_n(T)$ but not to $\Sigma_n(S)$, since otherwise $\Sigma_n(T)$ would contain $\bar{0}$. Hence, $\Sigma_n(S) \subsetneq \Sigma_n(T)$.
If $\Sigma_n(T)$ contains exactly one more element than $\Sigma_n(S)$, then this new element must be $\overline{\sigma(S) + x}$. However, $\bar{x} \in \Sigma_n(T)$, and $\bar{x} \ne \overline{\sigma(S) + x}$, so $\bar{x} \in \Sigma_n(S)$.
Let $d = \frac{n}{\text{gcd}(x,n)}$, where $d$ is the smallest positive integer such that $d\bar{x} = \bar{0}$. Suppose $\bar{x}, 2\bar{x}, \dots, p\bar{x} \in \Sigma_n(S)$, but $(p+1)\bar{x} \notin \Sigma_n(S)$. A set of the form
$$\{\bar{a}, \overline{a+x}, \overline{a+2x}, \dots, \overline{a+(d-1)x}\}$$
is called a coset of $\bar{x}$. The congruence classes modulo $n$ can be partitioned into $\frac{n}{d}$ cosets of $\bar{x}$.
If $\Sigma_n(S)$ contains some elements of a coset $C$ of $\bar{x}$ but not the entire coset, then $(\Sigma_n(S) \cap C) + \bar{x}$ will produce new elements not in $\Sigma_n(S) \cap C$. Since $\Sigma_n(T)$ contains only one more element than $\Sigma_n(S)$, it follows that $\Sigma_n(S)$ must consist of several complete cosets of $\bar{x}$ and one incomplete coset. This incomplete coset can only be $\{\bar{x}, 2\bar{x}, \dots, p\bar{x}\}$, and the new element in $\Sigma_n(T)$ is $(p+1)\bar{x}$, which must equal $\overline{\sigma(S)+x}$. Therefore, $\overline{\sigma(S)} = p\bar{x}$.
Step 2: Returning to the original problem, first add $n-1-\ell$ copies of 1 to $\{a_1, a_2, \dots, a_k\}$, obtaining a multiset $S$. It is easy to see that $m := |S| = k+(n-1)-\ell \ge \frac{n+1}{2}$ (using condition (1)), and
$$\Sigma_n(S) = \{\bar{1}, \bar{2}, \dots, \overline{n-1}\}.$$
Let the elements of $S$ be ordered as $b_1 \le b_2 \le \dots \le b_m$. Then $b_1 = 1$ (using condition (1), $\ell \le n-2$, so $S$ contains at least one 1), and $b_m < n$.
If for every $2 \le i \le m$, we have $b_i \le 1 + b_1 + \dots + b_{i-1}$, then by induction, it follows that
$$\Sigma(S) = \{1, 2, \dots, b_1 + b_2 + \dots + b_m\}.$$
Since $\Sigma_n(S) = \{\bar{1}, \bar{2}, \dots, \overline{n-1}\}$, it must be that $b_1 + b_2 + \dots + b_m = n-1$. Removing the added $n-1-\ell$ copies of 1, we obtain
$$a_1 + a_2 + \dots + a_k = \ell.$$
Now suppose there exists $2 \le i \le m$ (take the smallest such $i$) such that $b_i > 1 + b_1 + \dots + b_{i-1}$. Let $s = b_1 + \dots + b_{i-1} \ge i-1$. Then
$$\Sigma(\{b_1, \dots, b_{i-1}, b_i\}) = \{1, 2, \dots, s, b_i, b_i+1, \dots, b_i+s\}.$$
Since $b_i < n$, we have $b_i + s < n$ (otherwise $\overline{0} \in \Sigma_n(S)$), so
$$|\Sigma_n(\{b_1, \dots, b_i\})| = 2s + 1 \ge 2i - 1.$$
Now, iteratively add the remaining elements to $\{b_1, \dots, b_i\}$ such that at each step, $\Sigma_n$ gains at least two new elements, until no more can be added. Suppose we obtain $T \subset S$ with $|T| = t \ge i$, satisfying
$$|\Sigma_n(T)| \ge 2i - 1 + 2(t - i) = 2t - 1.$$
It follows that $t < m$, since otherwise $|\Sigma_n(T)| \ge 2m-1 \ge n$, which is a contradiction. Now, the remaining elements must each add only one new element to $\Sigma_n$ when included in $T$.
Claim: The remaining elements are all identical.
Suppose $x$ and $y$ are remaining elements, and both $\Sigma_n(T \cup \{x\})$ and $\Sigma_n(T \cup \{y\})$ contain exactly one more element than $\Sigma_n(T)$. By the conclusion of Step 1, $\Sigma_n(T)$ consists of several complete cosets of $\bar{x}$ and one incomplete coset $\{\bar{x}, 2\bar{x}, \dots, p\bar{x}\}$, with $\overline{\sigma(T)} = p\bar{x}$.
Since $\Sigma_n(T \cup \{y\})$ contains only one more element, $\overline{\sigma(T)+y} = \bar{y} + p\bar{x}$, it follows that $\bar{y}, \bar{y}+\bar{x}, \dots, \bar{y}+(p-1)\bar{x} \in \Sigma_n(T)$, but $\bar{y}+p\bar{x} \notin \Sigma_n(T)$. Therefore, $\bar{y}$ cannot belong to a complete coset of $\bar{x}$ in $\Sigma_n(T)$, so it must be in $\{\bar{x}, 2\bar{x}, \dots, p\bar{x}\}$. Consequently,
$$\{\bar{y}, \bar{y} + \bar{x}, \dots, \bar{y} + (p-1)\bar{x}\} \subset \{\bar{x}, 2\bar{x}, \dots, p\bar{x}\},$$
which implies $\bar{y} = \bar{x}$, i.e., $x = y$. The claim is proved.
Thus, the remaining elements are all equal to some $x$. Each time we add $x$, $\Sigma_n$ gains exactly one new element, successively adding $(p+1)\bar{x}, (p+2)\bar{x}, \dots, (d-1)\bar{x}$. In the final step, $S = T_1 \cup \{x\}$, and the new element added to $\Sigma_n$ is $(d-1)\bar{x} = \overline{\sigma(T_1)+x} = \overline{\sigma(S)}$, so $\bar{x} = -\overline{\sigma(S)}$.
Earlier, when we added 1 to $\{a_1, a_2, \dots, a_k\}$ to obtain $S$, each addition also introduced exactly one new element to $\Sigma_n$. Considering the last addition, $S = T_2 \cup \{1\}$, we similarly deduce $\overline{1} = -\overline{\sigma(S)}$, so $x = 1$. This means that in the previous process, the remaining elements were all 1. However, this is impossible because we had already taken $\{b_1, \dots, b_i\}$ with $b_i > 1$, which includes all the 1's. This contradiction completes the proof. $\square$