Let $t_{n}$ equal the integer closest to $\sqrt{n}$.

For example, $t_{1}=t_{2}=1$ since $\sqrt{1}=1$ and $\sqrt{2} \approx 1.41$ and $t_{3}=2$ since $\sqrt{3} \approx 1.73$.

The sum $\frac{1}{t_{1}}+\frac{1}{t_{2}}+\frac{1}{t_{3}}+\frac{1}{t_{4}}+\cdots+\frac{1}{t_{2008}}+\frac{1}{t_{2009}}+\frac{1}{t_{2010}}$ equals
(A) $88 \frac{1}{6}$
(B) $88 \frac{1}{2}$
(C) $88 \frac{2}{3}$
(D) $88 \frac{1}{3}$
(E) 90
First, we try a few values of $n$ to see if we can find a pattern in the values of $t_{n}$ :

| $n$ | $\sqrt{n}$ | $t_{n}$ | $n$ | $\sqrt{n}$ | $t_{n}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 1 | 1 | 11 | 3.32 | 3 |
| 2 | 1.41 | 1 | 12 | 3.46 | 3 |
| 3 | 1.73 | 2 | 13 | 3.61 | 4 |
| 4 | 2 | 2 | 14 | 3.74 | 4 |
| 5 | 2.24 | 2 | 15 | 3.87 | 4 |
| 6 | 2.45 | 2 | 16 | 4 | 4 |
| 7 | 2.65 | 3 | 17 | 4.12 | 4 |
| 8 | 2.83 | 3 | 18 | 4.24 | 4 |
| 9 | 3 | 3 | 19 | 4.36 | 4 |
| 10 | 3.16 | 3 | 20 | 4.47 | 4 |
|  |  |  | 21 | 4.58 | 5 |

(In each case, the $\sqrt{n}$ column contains an approximation of $\sqrt{n}$ to 2 decimal places.) 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.

ANSWER: (C)