Let [t] be the greatest integer less than or equal to t. Then the least value of $$p \in N$$ for which $$\lim_{x\rightarrow 0^{+}}\left(x([\frac{1}{x}]+[\frac{2}{x}]+...+[\frac{p}{x}])-x^{2}([\frac{1}{x^{2}}]+[\frac{2^{2}}{x^{2}}]+...+[\frac{9^{2}}{x^{2}}])\right) \geq 1$$ is equal to_______.

We use the identity $$[t] = t - \{t\}$$, where $$\{t\}$$ denotes the fractional part of $$t$$.

Evaluating the first sum:

$$x \sum_{k=1}^{p} \left[\frac{k}{x}\right] = x \sum_{k=1}^{p} \left(\frac{k}{x} - \left\{\frac{k}{x}\right\}\right) = \sum_{k=1}^{p} k - x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\}$$

$$= \frac{p(p+1)}{2} - x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\}$$

As $$x \to 0^+$$, since $$0 \leq \{t\} < 1$$ for all $$t$$, we have $$0 \leq x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\} < px$$.

Therefore $$\lim_{x \to 0^+} x \sum_{k=1}^{p} \left\{\frac{k}{x}\right\} = 0$$.

So the first part gives: $$\lim_{x \to 0^+} x \sum_{k=1}^{p} \left[\frac{k}{x}\right] = \frac{p(p+1)}{2}$$

Evaluating the second sum:

$$x^2 \sum_{k=1}^{9} \left[\frac{k^2}{x^2}\right] = x^2 \sum_{k=1}^{9} \left(\frac{k^2}{x^2} - \left\{\frac{k^2}{x^2}\right\}\right) = \sum_{k=1}^{9} k^2 - x^2 \sum_{k=1}^{9} \left\{\frac{k^2}{x^2}\right\}$$

Similarly, $$0 \leq x^2 \sum_{k=1}^{9} \left\{\frac{k^2}{x^2}\right\} < 9x^2 \to 0$$ as $$x \to 0^+$$.

So: $$\lim_{x \to 0^+} x^2 \sum_{k=1}^{9} \left[\frac{k^2}{x^2}\right] = \sum_{k=1}^{9} k^2 = \frac{9 \times 10 \times 19}{6} = 285$$

Combining:

The limit becomes: $$\frac{p(p+1)}{2} - 285 \geq 1$$

$$\frac{p(p+1)}{2} \geq 286$$

$$p(p+1) \geq 572$$

We check values: $$23 \times 24 = 552 < 572$$ and $$24 \times 25 = 600 \geq 572$$.

The least value of $$p \in \mathbb{N}$$ satisfying this inequality is $$p = 24$$.