For a real number $x,[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $\left[\frac{1}{2}\right]+\left[\frac{1}{2}+\frac{1}{100}\right]+\left[\frac{1}{2}+\frac{2}{100}\right]+\left[\frac{1}{2}+\frac{3}{100}\right]+\ldots \ldots \ldots \ldots .+\left[\frac{1}{2}+\frac{99}{100}\right]=$

The sum is $S = \left[\frac{1}{2}\right]+\left[\frac{1}{2}+\frac{1}{100}\right]+\cdots+\left[\frac{1}{2}+\frac{99}{100}\right]$.

Rewriting $\frac{1}{2}$ as $\frac{50}{100}$, each term becomes $\left[\frac{50+k}{100}\right]$ for $k = 0, 1, 2, \ldots, 99$.

When $\frac{50+k}{100} < 1$, which occurs for $k < 50$, the floor is $0$.
For $k = 0, 1, \ldots, 49$, the term equals $0$.

When $1 \leq \frac{50+k}{100} < 2$, which occurs for $k \geq 50$, the floor is $1$.
For $k = 50, 51, \ldots, 99$, the term equals $1$.

There are 50 terms in each group. Summing gives $S = 50 \times 0 + 50 \times 1 = 50$.

$\boxed{50}$