, 2021

$$U_n = (1 + 2 + 3 + \dots + (n-1)) \cdot n = \frac{(n-1)n^2}{2}.$$
Then $S_n = S_{n-1} + U_n$ for $n \ge 3$. Note that $U_n$ is divisible by 3 if $n \equiv 0 \text{ or } 1 \pmod 3$; and if $n \equiv 2 \pmod 3$, then $(n-1)n^2 \equiv 1 \pmod 3$ and is even, so $U_n \equiv 2 \pmod 3$. Hence $S_{n+3} \equiv S_n + 2 \pmod 3$ for $n \ge 2$. It is readily verified that $S_2 \equiv S_3 \equiv S_4 \equiv 2 \pmod 3$, so $S_5 \equiv S_6 \equiv S_7 \equiv 1 \pmod 3$ and $S_8 \equiv S_9 \equiv S_{10} \equiv 0 \pmod 3$, and it follows that $S_n$ is divisible by 3 if and only if $n \equiv 0 \text{ or } \pm 1 \pmod 9$. Thus the sum of the 10 least values of $n$ that satisfy the required condition is
$$8 + 9 + 10 + 17 + 18 + 19 + 26 + 27 + 28 + 35 = 197.$$

The sum of the products $jk$ as $j$ and $k$ run independently from 1 to $n$ is
$$(1 + 2 + \dots + n)^2 = \left(\frac{n(n+1)}{2}\right)^2.$$
To eliminate the cases in which $j = k$, subtract
$$1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
Thus
$$\sum_{\substack{1 \le j \le n \\ 1 \le k \le n \\ j \ne k}} jk = \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6} = \frac{(n-1)n(n+1)(3n+2)}{12}.$$
For a given pair $j, k$ with $j \ne k$ either $j < k$ or $j > k$, but their product is the same in either order. To impose the condition $j < k$, it suffices to divide by 2. Thus
$$S_n = \frac{(n-1)n(n+1)(3n+2)}{24}.$$
There is one factor of 3 in the denominator. For any $n$, exactly one of $n+1, n, n-1$ is divisible by 3, and $3n+2$ is not divisible by 3. In order that $S_n$ be divisible by 3 it is necessary and sufficient that the factor that is divisible by 3 should in fact be divisible by 9. That is, $n \equiv 0 \text{ or } \pm 1 \pmod 9$, and the answer can be calculated as above.