Let $a_n$ be the closest to $\sqrt n$ integer. 
Find the sum $1/a_1 + 1/a_2 + ... + 1/a_{1980}$.
1. We start by understanding the sequence \(a_n\), which is defined as the closest integer to \(\sqrt{n}\). For any integer \(m\), the values of \(a_n\) change at the squares of integers. Specifically, \(a_n = m\) for \(m^2 \leq n < (m+1)^2\).

2. We need to find the sum \( \sum_{k=1}^{1980} \frac{1}{a_k} \).

3. Observe that for each \(m\), the integer \(a_n\) remains constant for \(2m+1\) values of \(n\) (from \(m^2\) to \((m+1)^2 - 1\)). However, since \(a_n\) is the closest integer to \(\sqrt{n}\), we need to adjust the range slightly:
   - For \(n = m^2 + m\), \(a_n = m\).
   - For \(n = m^2 + m + 1\) to \((m+1)^2 + (m+1)\), \(a_n = m + 1\).

4. Therefore, for each \(m\), there are \(2m + 2\) terms where \(a_n = m + 1\).

5. We need to find the largest \(m\) such that \((m+1)^2 + (m+1) \leq 1980\):
   - Solving \((m+1)^2 + (m+1) = 1980\) gives \(m = 43\).

6. Now, we sum the reciprocals of \(a_n\) for \(n\) from 1 to 1980:
   \[
   \sum_{m=0}^{43} \sum_{k=m^2+m+1}^{(m+1)^2+(m+1)} \frac{1}{a_k} = \sum_{m=0}^{43} (2m+2) \cdot \frac{1}{m+1}
   \]

7. Simplifying the inner sum:
   \[
   \sum_{m=0}^{43} (2m+2) \cdot \frac{1}{m+1} = \sum_{m=0}^{43} 2 = 44 \cdot 2 = 88
   \]

8. Therefore, the sum \( \sum_{k=1}^{1980} \frac{1}{a_k} \) is 88.

The final answer is \(\boxed{88}\).