1. We start by considering the polynomial $(1+x)^n$. The sum of the squares of the coefficients of this polynomial is given by $S_n$.

2. We need to prove that $S_{2n} + 1$ is not divisible by 3.

3. By definition, $S_n$ is the sum of the squares of the coefficients of $(1+x)^n$. The coefficients of $(1+x)^n$ are given by the binomial coefficients $\binom{n}{k}$ for $k = 0, 1, \ldots, n$. Therefore,
$$S_n = \sum_{k=0}^{n} \binom{n}{k}^2.$$

4. Using the identity for the sum of the squares of binomial coefficients, we have:
$$S_n = \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}.$$

5. Therefore, $S_{2n} = \binom{4n}{2n}$. We need to show that $S_{2n} + 1 = \binom{4n}{2n} + 1$ is not divisible by 3.

6. We use a lemma related to binomial coefficients modulo a prime $p$:
$$\text{Lemma: If } \binom{2x}{x} \text{ is not divisible by } p, \text{ and } x = (a_0, a_1, \ldots, a_t)_p \text{ (base } p \text{ representation)}, \text{ then } 0 \leq a_i \leq \frac{p-1}{2}.$$

7. Applying this lemma with $p = 3$, we consider $2n$ in base 3 representation. Let $2n = (a_0, a_1, \ldots, a_t)_3$ where $a_i \in \{0, 1\}$.

8. By the lemma, if $\binom{4n}{2n}$ is not divisible by 3, then the sum of the digits $a_i$ in the base 3 representation of $2n$ must be even. This is because each $a_i$ can only be 0 or 1, and the sum of an even number of 1's is even.

9. Suppose for contradiction that $\binom{4n}{2n} \equiv -1 \pmod{3}$. Then:
$$\binom{4n}{2n} = \sum_{i=0}^{t} \binom{2a_i}{a_i} \pmod{3}.$$
Notice that each term $\binom{2a_i}{a_i}$ is 1 if $a_i = 0$ and -1 if $a_i = 1$.

10. For $\binom{4n}{2n}$ to be $-1 \pmod{3}$, there must be an odd number of $a_i$ equal to 1. However, this implies that the sum of the digits $a_i$ in the base 3 representation of $2n$ is odd.

11. But $2n = a_0 + a_1 \cdot 3 + \ldots + a_t \cdot 3^t$ is even, which contradicts the fact that the sum of the digits $a_i$ is odd.

12. Therefore, $\binom{4n}{2n} \equiv 1 \pmod{3}$, and hence $S_{2n} + 1 = \binom{4n}{2n} + 1 \equiv 2 \pmod{3}$, which is not divisible by 3.

$\blacksquare$