First, we try a few values of $n$ to see if we can find a pattern in the values of $t_{n}$: So $t_{n}=1$ for 2 values of $n, 2$ for 4 values of $n, 3$ for 6 values of $n, 4$ for 8 values of $n$. We conjecture that $t_{n}=k$ for $2 k$ values of $n$. We will prove this fact at the end of the solution. Next, we note that $\sqrt{2010} \approx 44.83$ and so $t_{2010}=45$. This means that before this point in the sequence, we have included all terms with $t_{n} \leq 44$. According to our conjecture, the number of terms with $t_{n} \leq 44$ should be $2+4+6+\cdots+86+88=2(1+2+3+\cdots+43+44)=2\left(\frac{1}{2}(44)(45)\right)=44(45)=1980$. Note that $\sqrt{1980} \approx 44.497$ and $\sqrt{1981} \approx 44.508$ so $t_{1980}=44$ and $t_{1981}=45$. Since $t_{1981}=t_{2010}=45$, then each of the terms from $t_{1981}$ to $t_{2010}$ equals 45. Therefore, there are 30 terms that equal 45. Thus, the required sum equals $2\left(\frac{1}{1}\right)+4\left(\frac{1}{2}\right)+6\left(\frac{1}{3}\right)+\cdots+86\left(\frac{1}{43}\right)+88\left(\frac{1}{44}\right)+30\left(\frac{1}{45}\right)=2+2+2+\cdots+2+2+\frac{2}{3}$ where there are 44 copies of 2. Therefore, the sum equals $88 \frac{2}{3}$. Lastly, we prove that for each positive integer $k$, there are $2 k$ terms $t_{n}$ that equal $k$: In order to have $t_{n}=k$, we need $k-\frac{1}{2} \leq \sqrt{n}<k+\frac{1}{2}$ (in other words, $\sqrt{n}$ needs to round to $k$). Since $n$ and $k$ are positive, then $k-\frac{1}{2} \leq \sqrt{n}$ is equivalent to $\left(k-\frac{1}{2}\right)^{2} \leq n$ and $\sqrt{n}<k+\frac{1}{2}$ is equivalent to $n<\left(k+\frac{1}{2}\right)^{2}$. Therefore, we need $\left(k-\frac{1}{2}\right)^{2} \leq n<\left(k+\frac{1}{2}\right)^{2}$ or $k^{2}-k+\frac{1}{4} \leq n<k^{2}+k+\frac{1}{4}$. Since $n$ is an integer, then $k^{2}-k+1 \leq n \leq k^{2}+k$. There are thus $\left(k^{2}+k\right)-\left(k^{2}-k+1\right)+1=2 k$ such values of $n$, as required.