Do you like sums?

Let's generalize!!! Thanks to @Emil Gueorguiev for pointing out I could further generalize by changing every instance of $10^n$ in my original solution to $n$ . The following is the result of making this change (which helps readability too since I no longer have exponents in my summation indices)

Let $r$ be any real number and let $n$ be some non-negative integer. Then the sum we're interested in is $\begin{aligned} \sum_{k=0}^{n-1} \left\lfloor r + \frac{k}{n} \right\rfloor &= \sum_{k=0}^{n-1} \left\lfloor \lfloor r\rfloor + \{r\} + \frac{k}{n} \right\rfloor \\ &= \sum_{k=0}^{n-1} \lfloor r \rfloor + \left\lfloor \frac{n\{r\} + k}{n} \right\rfloor \\ &= n\lfloor r \rfloor + \sum_{k=0}^{n-1} \left\lfloor \frac{\left\lfloor n\{r\}\right\rfloor + k}{n} \right\rfloor \\ &= n\lfloor r \rfloor + \sum_{k=0}^{n - \left\lfloor n\{r\}\right\rfloor - 1} 0 + \sum_{k=n - \left\lfloor n\{r\}\right\rfloor}^{n - 1} 1 \\ &= n\lfloor r \rfloor + \left\lfloor n\{r\}\right\rfloor \\ &= \Big\lfloor n \lfloor r \rfloor + n\{r\}\Big\rfloor \\ &= \lfloor n r\rfloor \end{aligned}$

Set $r=\sqrt{2}$ and $n=100$ to make our sum identical to the one in the problem, giving an answer of $\left\lfloor 100 \sqrt{2}\right\rfloor = \lfloor 141.421356 \rfloor = \boxed{141}$

The general term of the given sum can be written as $\left \lfloor \sqrt 2+\dfrac k{100}\right \rfloor$ , where $k$ is an integer and $0\le k \le 99$ . Since $\sqrt 2+\dfrac 99{100} \approx 2.404 < 3$ ,

$\left \lfloor \sqrt 2+\dfrac k{100}\right \rfloor= \begin{cases} 1 & \text{if } \sqrt 2+\dfrac k{100} < 2 \text{ or }0 \le k \le 58 \\ 2 & \text{if } \sqrt 2+\dfrac k{100} \ge 2 \text{ or } 59 \le k \le 99 \end{cases}$

Then we have:

$\begin{aligned} S & = \sum_{k=0}^{99} \left \lfloor \sqrt 2 + \frac k{100} \right \rfloor = \sum_{k=0}^{58} 1 + \sum_{k=59}^{99} 2 = 59 + 82 = \boxed{141} \end{aligned}$