7. Let $[x]$ denote the greatest integer not exceeding $x$, then $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{99}]$, find the value of $[\sqrt{S}]$.

Let's translate the problem and solution step by step:

1. **Understanding the Notation:**
   - $[x]$ represents the greatest integer less than or equal to $x$.

2. **Expression for $S$:**
   - $S = [\sqrt{1}] + [\sqrt{2}] + [\sqrt{3}] + \cdots + [\sqrt{99}]$.

3. **Breaking Down the Sum:**
   - We need to evaluate the sum of the greatest integer parts of the square roots from 1 to 99.

4. **Grouping the Terms:**
   - For $1 \leq k \leq 3$, $[\sqrt{k}] = 1$.
   - For $4 \leq k \leq 8$, $[\sqrt{k}] = 2$.
   - For $9 \leq k \leq 15$, $[\sqrt{k}] = 3$.
   - For $16 \leq k \leq 24$, $[\sqrt{k}] = 4$.
   - For $25 \leq k \leq 35$, $[\sqrt{k}] = 5$.
   - For $36 \leq k \leq 48$, $[\sqrt{k}] = 6$.
   - For $49 \leq k \leq 63$, $[\sqrt{k}] = 7$.
   - For $64 \leq k \leq 80$, $[\sqrt{k}] = 8$.
   - For $81 \leq k \leq 99$, $[\sqrt{k}] = 9$.

5. **Calculating the Sum:**
   - Sum for $1 \leq k \leq 3$: $3 \times 1 = 3$.
   - Sum for $4 \leq k \leq 8$: $5 \times 2 = 10$.
   - Sum for $9 \leq k \leq 15$: $7 \times 3 = 21$.
   - Sum for $16 \leq k \leq 24$: $9 \times 4 = 36$.
   - Sum for $25 \leq k \leq 35$: $11 \times 5 = 55$.
   - Sum for $36 \leq k \leq 48$: $13 \times 6 = 78$.
   - Sum for $49 \leq k \leq 63$: $15 \times 7 = 105$.
   - Sum for $64 \leq k \leq 80$: $17 \times 8 = 136$.
   - Sum for $81 \leq k \leq 99$: $19 \times 9 = 171$.

6. **Total Sum $S$:**
   - $S = 3 + 10 + 21 + 36 + 55 + 78 + 105 + 136 + 171 = 615$.

7. **Finding $[\sqrt{S}]$:**
   - $S = 615$.
   - $\sqrt{615} \approx 24.8$.
   - $[\sqrt{615}] = 24$.

Therefore, the value of $[\sqrt{S}]$ is $\boxed{24}$.
For $k^{2} \leqslant i \leqslant k^{2}+2 k$, $[\sqrt{i}]=k, k=1,2, \cdots, 9$, thus $S=\sum_{i=1}^{99}[\sqrt{i}]=\sum_{k=1}^{9} k(2 k+1)=2 \sum_{k=1}^{9} k^{2}+\sum_{k=1}^{9} k=\frac{9 \cdot 10 \cdot 19}{3}+\frac{9 \cdot 10}{2}=615$. Therefore, $[\sqrt{S}]=[\sqrt{615}]=24$.