Let $S_1(x)=x\sum_{k=1}^{p}\left[\frac{k}{x}\right],\qquad S_2(x)=x^2\sum_{k=1}^{9}\left[\frac{k^2}{x^2}\right].$ We need the least $p\in \mathbb N$ such that $\lim_{x\to 0^+}\left(S_1(x)-S_2(x)\right)\ge 1.$

1. Use the standard floor estimate

For any real $t$, $t-1<[t]\le t.$

Hence, for fixed positive integer $k$, $\frac{k}{x}-1<\left[\frac{k}{x}\right]\le \frac{k}{x}.$ Multiplying by $x>0$, $k-x<x\left[\frac{k}{x}\right]\le k.$ As $x\to 0^+$, $x\left[\frac{k}{x}\right]\to k.$

Similarly, $\frac{k^2}{x^2}-1<\left[\frac{k^2}{x^2}\right]\le \frac{k^2}{x^2},$ so multiplying by $x^2$, $k^2-x^2<x^2\left[\frac{k^2}{x^2}\right]\le k^2.$ Thus, $x^2\left[\frac{k^2}{x^2}\right]\to k^2 \quad (x\to 0^+).$

Therefore, $\lim_{x\to 0^+}S_1(x)=\sum_{k=1}^{p}k=\frac{p(p+1)}{2},$ અને $\lim_{x\to 0^+}S_2(x)=\sum_{k=1}^{9}k^2=\frac{9\cdot 10\cdot 19}{6}=285.$

So, $\lim_{x\to 0^+}(S_1(x)-S_2(x))=\frac{p(p+1)}{2}-285.$

2. Impose the required condition

We need $\frac{p(p+1)}{2}-285\ge 1.$ So, $\frac{p(p+1)}{2}\ge 286,$ which gives $p(p+1)\ge 572.$

Now check nearby integers:

• For $p=23$: $23\cdot 24=552<572$
• For $p=24$: $24\cdot 25=600\ge 572$

Hence the least such natural number is $p=24.$

3. Comparison with stored answer

Stored correct answer: $24$

Our derived answer matches it.