Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\mathbb{N}$ denote the set of all positive integers.

Suppose, to the contrary, that there exist only finitely many positive integers that are not clean. Therefore, there exists a positive integer $N$ such that every integer $n > N$ has exactly one odd representation.

Clearly, in this case, $S$ must be infinite. We first prove the following properties of odd and even representations.

1. **Any positive integer $n$ has at most one odd and at most one even representation.**

Proof. We first show that every integer $n$ has at most one even representation. Since $S$ is infinite, there exists $x \in S$ such that $x > \max\{n, N\}$. Then, the number $n+x$ must be clean, and $x$ does not appear in any even representation of $n$ (since $x > n$). If $n$ has more than one even representation, then we obtain two distinct odd representations of $n+x$ by adding $x$ to the even representations of $n$, which is a contradiction. Hence, $n$ has at most one even representation.

Similarly, there exist two elements $y, z \in S$ such that $y, z > \max\{n, N\}$.
If $n$ has more than one odd representation, then $n+y+z$ has multiple odd representations, which is a contradiction.

2. **Fix $s \in S$. Suppose that a number $n > N$ has no even representation. Then $n+2s$ has an even representation containing $s$ for all integers $a \ge 1$.**

Proof. It is sufficient to prove the following statement: If $n$ has no even representation without $s$, then $n+2s$ has an even representation containing $s$ (and hence no even representation without $s$ by Property 1.)

Since $n + s > n > N$ is clean, it has an odd representation. In addition, notice that the odd representation of $n+s$ does not contain $s$; otherwise, $n$ has an even representation without $s$, which is a contradiction. Hence, by adding $s$ to the representation, we get an even representation of $n+2s$ containing $s$.

3. Every sufficiently large integer has an even representation.

Proof. Fix any $s \in S$, and let $r$ be an arbitrary element in $\{1, 2, \dots, 2s\}$. Then, Property 2 implies that the set $\mathbb{N}_r = \{r + 2as : a \ge 0\}$ contains at most one number exceeding $N$ with no even representation. Hence, $\mathbb{N}_r$ contains finitely many positive integers with no even representation, and so does $\mathbb{N} = \bigcup_{r=1}^{2s} \mathbb{N}_r$.

Combining Properties 1-3, we may assume that $N$ is chosen such that $n > N$ has exactly one odd and exactly one even representation. In particular, each element $s > N$ of $S$ has an even representation.

4. **For any $s, t \in S$ with $N < s < t$, the even representation of $t$ contains $s$.**

Proof. Suppose NOT. Then, $s+t$ would have two odd representations (one by adding $t$ to the even representation of $s$, and the other by adding $s$ to the even representation of $t$), which contradicts Property 1.

We are now ready for the original problem. Let $s_1 < s_2 < \dots$ be all elements of $S$, and set $\sigma_n = \sum_{i=1}^n s_i$ for each nonnegative integer $n$. Fix an integer $k$ such that $s_k > N$. Then, Property 4 implies that for every $i > k$, the even representation of $s_i$ contains all the numbers $s_k, s_{k+1}, \dots, s_{i-1}$. Therefore,

$$s_i = s_k + s_{k+1} + \dots + s_{i-1} + R_i = \sigma_{i-1} - \sigma_{k-1} + R_i, \quad (1)$$

where $R_i$ is a sum of some of $s_1, s_2, \dots, s_{k-1}$. In particular, $0 \le R_i \le s_1 + s_2 + \dots + s_{k-1} = \sigma_{k-1}$.

Let $j_0$ be an integer satisfying $j_0 > k$ and $\sigma_{j_0} > 2\sigma_{k-1}$. Then (1) shows that, for every $j > j_0$,

$$s_{j+1} \ge \sigma_j - \sigma_{k-1} > \sigma_j/2. \quad (2)$$

Next, let $p > j_0$ be an index such that $R_p = \min_{i>j_0} R_i$. Then,

$$s_{p+1} = s_k + s_{k+1} + \dots + s_p + R_{p+1} = (s_p - R_p) + s_p + R_{p+1} \ge 2s_p.$$

Therefore, there is no element of $S$ larger than $s_p$ but smaller than $2s_p$. It follows that the even representation $\tau$ of $2s_p$ does not contain any element larger than $s_p$. On the other hand, inequality (2) yields $2s_p > s_1 + s_2 + \dots + s_{p-1}$, so $\tau$ must contain a term larger than $s_{p-1}$. Thus, it must contain $s_p$. After removing $s_p$ from $\tau$, we have that $s_p$ has an odd representation not containing $s_p$, which contradicts Property 1 since $s_p$ itself also forms an odd representation of $s_p$.