The number of distinct integers $n$ for which $\log_{1/4}(n^2 - 7n + 11) > 0$, is

Since the base $\frac{1}{4}$ is less than 1, the inequality $\log_{1/4}(n^2 - 7n + 11) > 0$ holds only when:

$0 < n^2 - 7n + 11 < 1$

This is because $\left(\dfrac{1}{4}\right)^0 = 1$, and raising $\dfrac{1}{4}$ to a positive power gives a value less than 1. So the logarithm is positive only when the argument is strictly between 0 and 1.

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Solving the right inequality:

$n^2 - 7n + 11 < 1$

$n^2 - 7n + 10 < 0$

$(n - 2)(n - 5) < 0$

This holds when $2 < n < 5$, so the integer candidates are $n = 3$ and $n = 4$.

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Now checking the left inequality $n^2 - 7n + 11 > 0$ for these candidates:

For $n = 3$:

$9 - 21 + 11 = -1$

This is negative, so it does not satisfy $n^2 - 7n + 11 > 0$.

For $n = 4$:

$16 - 28 + 11 = -1$

This is also negative, so it does not satisfy the condition either.

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Checking the boundary values $n = 2$ and $n = 5$:

For $n = 2$: $\quad 4 - 14 + 11 = 1 \quad \Rightarrow \quad \log_{1/4}(1) = 0$

For $n = 5$: $\quad 25 - 35 + 11 = 1 \quad \Rightarrow \quad \log_{1/4}(1) = 0$

Both give a logarithm value of $0$, not strictly greater than $0$.

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A summary of nearby integer values:

$n$	$n^2 - 7n + 11$	Lies in $(0,1)$?
$1$	$5$	No
$2$	$1$	No
$3$	$-1$	No
$4$	$-1$	No
$5$	$1$	No
$6$	$5$	No

No integer value of $n$ makes $n^2 - 7n + 11$ lie strictly between $0$ and $1$.

The answer is $\boxed{0}$.