Finite Population Variance for a Changing Population How does the addition of one unit affect the population variance of a finite population if everything else remains unchanged? What are the conditions such that the new unit leaves the variance unchanged (increases/decreases it)?
I was able to find the following paper regarding sample variances for changing finite populations:
http://www.amstat.org/sections/srms/Proceedings/papers/1987_087.pdf.
But I am asking specifically about population variances. Any help is appreciated.

I was unable to find the sample calculations that correspond to the specific problem here (as suggested by Glen_b), but I was able to confirm the following answer with numerical calculations in R at the bottom of this answer.
Let $N$ be the initial number of units in the population and $N + 1$ be the number of units in the population after the change. Denote the initial set of observations $X = \{x_1, \ldots, x_N\}$ (i.e., one observation corresponding to each population unit). Denote the set of observations after the change $Y = X \cup \{x_{N+1}\}$.
The mean of $X$ is
$\mu_X = \frac{\sum_{i=1}^N{x_i}}{N}$.
The mean of Y is
$\mu_Y = \frac{\sum_{i=1}^{N+1}{x_i}}{N+1}
= \mu_X \frac{N}{N+1} + \frac{x_{N+1}}{N+1}$
Define $x_{N+1}$ as the original mean, $\mu_X$, plus some $\varepsilon$. Then,
the mean of $Y$ is
$\mu_Y = \mu_X \frac{N}{N+1} + \frac{\mu_X
+ \varepsilon}{N+1} = \mu_X + \frac{\varepsilon}{N+1}$
The variance of $Y$ is 
$\sigma^2_Y = \frac{\sum_{i=1}^{N+1}
\left(x_i - \mu_Y \right)^2}{N+1} =
\frac{\sum_{i=1}^{N+1}
\left(x_i - \mu_X - \frac{\varepsilon}{N + 1} \right)^2}{N+1}$
$= \frac{\sum_{i=1}^{N} x_i^2 + \mu_X^2
+ \frac{\varepsilon^2}{\left(N+1\right)^2} - 2x_i\mu_X
- 2x_i\frac{\varepsilon}{N+1} + 2\mu_X\frac{\varepsilon}{N+1}}{N + 1}$
$\frac{\left(\mu_X + \varepsilon - \mu_X - \frac{\varepsilon}{N + 1}\right)}{N
+ 1} $
$ = \frac{N}{N+1}\sigma^2_X + \frac{N\varepsilon^2}{\left(N+1\right)^3}
    - \frac{2N\mu_X\varepsilon}{\left(N+1\right)^2}
    + \frac{2N\mu_X\varepsilon}{\left(N+1\right)^2}
    + \frac{N^2\varepsilon^2}{\left(N+1\right)^3}$
$ = \frac{N}{N+1}
\sigma^2_X + \frac{N}{\left(N+1\right)^2}\varepsilon^2$
When $x_{N+1}$ is equal to $\mu_X$, the variance of $Y$ is 
$\frac{N}{N+1}\sigma^2_X < \sigma^2_X $
Thus, when $\varepsilon$ is sufficiently small $\sigma^2_Y$ is less than
$\sigma^2_X$. To determine how large $\varepsilon$ should be so that the
variance of $Y$ is greater than the variance of $X$, I set the two variances
equal.
$ \frac{N}{N+1} \sigma^2_X
    + \frac{N}{\left(N+1\right)^2}\varepsilon^2 = \sigma^2_X$
$   \frac{N}{\left(N+1\right)^2}\varepsilon^2 = \frac{1}{N+1} \sigma^2_X$
$   \varepsilon^2 = \frac{N+1}{N} \sigma^2_X $
$ \varepsilon = \pm \sigma_X
    \sqrt{\frac{N+1}{N}}$
Thus, adding a unit who's observation is within $\sqrt{\frac{N+1}{N}}$ standard deviations
of the old mean will lead to a lower variance.

The following R script verifies the above conclusion:
N <- 10
X <- runif(N)
width <- sqrt((N+1)/N)
# on the boundary
var(c(X, mean(X) + width * sqrt(var(X)))) - var(X) == 0
# outside the boundary
var(c(X, mean(X) + width * sqrt(var(X)) + 1)) - var(X) > 0
# inside the boundary
var(c(X, mean(X))) - var(X) < 0

