The answer is $10$, achieved when $S_i = \{ i\}$ for $i < 10$ and $S_10 = \{10, 11, \dots, 100 \}.$ The key insight is the following: Claim: Fix $n$ for the moment. Then the minimum score is obtained when $S_i = \{ i \}$ for $i < n$ and $S_n = \{n, n+1, \dots, 100 \}$. The idea is that we can perform swaps that decrease the score until we reach that partition. Suppose we have another distinct partition. Let $f(i)$ denote the size of the $S_j$ which contains $i$. Let $k$ be the minimal integer such that $f(k) > 1$. Then suppose $k$ belongs to a set of the form $\{ k \} \cup A $. Because $A \neq \{ k+1, \dots 100 \}$, there must exist another set $B$ among the $S_i$ such that every element of $B$ is at least $k$. Then we claim that isolating $\{k \}$ and combining $A$ and $B$ into a single set decreases the score. This is a simple computation. Denote the score after this swap by $S'$. Let $\Sigma{T}$ denote the sum of elements of an arbitrary set $T$. Then \begin{align*} n(S - S') &= \frac{ k + \sum(A)}{|A| + 1} + \frac{\sum(B)}{|B|} - k - \frac{\sum (A) + \sum (B)}{|A| + |B|} \\ &= \frac{\sum(A) |B| (|B|-1) + \sum(B) |A|(|A| + 1) - k |A| |B| (|A|+|B|)}{(|A|+1)|B|(|A|+|B|)} \\ \end{align*}By the minimality of $k$, we have $\sum(A) > k|A|$ and likewise $\sum(B) > k|B|. $ Thus \[ n(S - S') > \frac{ k|A||B|(|B|-1 + |A| + 1|) - k|A||B|(|A|+|B|)}{(|A|+1)|B|(|A|+|B|)} > 0 . \]Hence we may perform swaps that decrease the score unless our partition is of the aforementioned form. This proves the claim. Thus we are left with optimising a function of $n$. For fixed $n$, the minimal partition has score \[ \frac{1 + 2 + \dots + n-1 + \frac{100+n}{2}}{n} = \frac{n^2 + 100}{2n} = \frac{n}{2} + \frac{50}{n}. \]By AM-GM this is at most 10.

Consider $S = a_1+a_2+...+a_n$ and denote $s_i = |S_i|$ : we have $S = \sum_{i=1}^{n} \sum_{s \in S_i} \frac{1}{s_i} s$ Now look at sequences $(1,2,..., 100)$ and $(\frac{1}{s_1}, \frac{1}{s_1}, ..., \frac{1}{s_2}, ... ,\frac{1}{s_n})$. Rearrangement inequality tells us that our sum is greater then or equal then the case where the smallest $\frac{1}{s_i}$'s are paired with the biggest numbers and vice versa. But in that case the numbers $1$ through $100$ are partitioned in that order in these sets (sorry for the dumb explanations to whoever reads this). Now we set $s_0 = 0$ and $b_j = \sum_{i=1}^{j} s_i$ for convenience and write the new sum as $S \geq \sum_{i=1}^{n} \frac{(b_{i-1}+1)+(b_{i-1}+2)+...+(b_{i-1}+s_{i})}{s_i} = \sum_{i=1}^{n} (b_{i-1} + \frac{s_i+1}{2}) = b_1+b_2+...+b_{n-1} + \frac{n}{2} + 50$ Note that the last equality follows from $\sum_{i=1}^{n} s_i = 100$. Obviously $b_i \geq i$, so $b_1+...+b_{n-1} \geq 1+...+(n-1) =\frac{n^2-n}{2}$, so we get $S \geq 50+ \frac{n^2}{2}$. Then we have the final inequality $\frac{a_1+...+a_n}{n} = \frac{S}{n} \geq \frac{50}{n} + \frac{n}{2} \geq 2\sqrt{25} = 10$ from $AM-GM$. All of the above inequalities motivate us to construct an example the same way @Blastzit did in his solution

The answer is $10$. We can achieve this with the partitions $\{1 \},$ $\{2 \}, ...$ $\{9 \},$ $\{10,11, \dots 100 \}$ Lemma: If there exists indices $i$ and $j$ such that $|S_i|<|S_j|$, and there exists $k\in S_i$, $l\in S_j$, $k>j$, then the partition is not optimal. In particular, an optimal partition will not have such elements. Proof: By swapping $k$ and $j$, the sum ${a_1+a_2+\ldots+a_n}$ decreases by $\frac{k-l}{|S_i|} - \frac{k-l}{|S_j|} = \frac{(k-l)(|S_j|-|S_i|)}{|S_j||S_i|}>0$, hence this is suboptimal since our score decreased. Now observe that in any optimal partition, if we order the sets such that the sets are in order of increasing sizes and we impose the additional constraint that if $|S_i|=|S_j|$, $i<j$, then the minimum element of $S_i$ is smaller than the minimum element of $S_j$, we note that writing the elements of $S_1$, then $S_2$, etc. in increasing order yields the sequence $1,2,\dots n$ In particular, for some $r_i$ with $1\le r_1 \le r_2 \le \dots r_{n-1}$, we have: $S_1 = \{1,2,\dots,r_1 \}$, $S_2= \{r_1+1,r_1+2,\dots r_2 \}$ $\dots$ $S_n= \{r_{n-1}+1, r_{n-1}+2, \dots 100 \}$ Then the score can be simply expressed as $Score = \frac{\frac{1+r_1}{2} + \frac{(r_1+1)+r_2}{2} + \dots + \frac{(r_{n-1}+1)+100}{2}}{n} = \frac{1}{2}(\frac{100+n+2\Sigma r_i}{n})$ Furthermore, we know that $r_i\ge i$, hence this summation is at least $\frac{1}{2}(\frac{100+n+(n)(n-1)}{n})=\frac{1}{2} + \frac{1}{2}(\frac{100}{n} + (n-1))$ which we know by AM-GM to be minimised when we have $n=10$, so that the score becomes at least $\frac{1}{2} + \frac{1}{2}(\frac{100}{10} + 9)=10$ as desired.

Excellent!

HKIS200543 wrote: The answer is $10$, achieved when $S_i = \{ i\}$ for $i < 10$ and $S_10 = \{10, 11, \dots, 100 \}.$ The key insight is the following: Claim: Fix $n$ for the moment. Then the minimum score is obtained when $S_i = \{ i \}$ for $i < n$ and $S_n = \{n, n+1, \dots, 100 \}$. The idea is that we can perform swaps that decrease the score until we reach that partition. Suppose we have another distinct partition. Let $f(i)$ denote the size of the $S_j$ which contains $i$. Let $k$ be the minimal integer such that $f(k) > 1$. Then suppose $k$ belongs to a set of the form $\{ k \} \cup A $. Because $A \neq \{ k+1, \dots 100 \}$, there must exist another set $B$ among the $S_i$ such that every element of $B$ is at least $k$. Then we claim that isolating $\{k \}$ and combining $A$ and $B$ into a single set decreases the score. This is a simple computation. Denote the score after this swap by $S'$. Let $\Sigma{T}$ denote the sum of elements of an arbitrary set $T$. Then \begin{align*} n(S - S') &= \frac{ k + \sum(A)}{|A| + 1} + \frac{\sum(B)}{|B|} - k - \frac{\sum (A) + \sum (B)}{|A| + |B|} \\ &= \frac{\sum(A) |B| (|B|-1) + \sum(B) |A|(|A| + 1) - k |A| |B| (|A|+|B|)}{(|A|+1)|B|(|A|+|B|)} \\ \end{align*}By the minimality of $k$, we have $\sum(A) > k|A|$ and likewise $\sum(B) > k|B|. $ Thus \[ n(S - S') > \frac{ k|A||B|(|B|-1 + |A| + 1|) - k|A||B|(|A|+|B|)}{(|A|+1)|B|(|A|+|B|)} > 0 . \]Hence we may perform swaps that decrease the score unless our partition is of the aforementioned form. This proves the claim. Thus we are left with optimising a function of $n$. For fixed $n$, the minimal partition has score \[ \frac{1 + 2 + \dots + n-1 + \frac{100+n}{2}}{n} = \frac{n^2 + 100}{2n} = \frac{n}{2} + \frac{50}{n}. \]By AM-GM this is at most 10.