When n increases the t-value increases in a hypothesis test, but the t-table is just the opposite. Why? The formula for $t$ in a hypothesis test is given by:
$$
t=\frac{\bar{X}-\mu}{\hat \sigma/\sqrt{n}}. 
$$
When $n$ increases, the $t$-value increases according to the above formula. But why does critical $t$-value decrease in the $t$-table as $\text{df}$ (which is a function of $n$) increases?
 A: These are two different phenomena:  


*

*$t$-statistic
Holding all else constant, if $N$ increases the $t$-value must increase as a simple matter of arithmetic.  Consider the fraction in the denominator, $\hat\sigma/\sqrt{n}$, if $n$ gets bigger, then $\sqrt n$ will get bigger as well (albeit more slowly), because the square root is a monotonic transformation.  Since the square root of $n$ is the denominator of that fraction, as it gets bigger, the fraction will get smaller.  However, this fraction is, in turn, a denominator.  As a result, as that denominator gets smaller, the second fraction gets bigger.  Thus, the $t$-value will get bigger as $n$ gets bigger.  (Assuming, again, that $\hat\sigma$ and $(\bar x - \mu_{\rm null})$ remain the same.)  
What does this mean conceptually?  Well, the more data we have / the closer the sample size gets to the population size, the less far the sample mean will tend to vary from the population mean due to sampling error (cf., the law of large numbers).  With a small, finite population, this is easy to see, but although it may not be as intuitive, the same holds true if the population is infinite.  Since the sample mean ($\bar x$) shouldn't fluctuate very far from the reference (null) value, we can be more confident that the observed distance of the sample mean from the null is because the null value is not actually the mean of the population from which the sample was drawn.  More accurately, it becomes less and less probable to have found a sample mean that far or further away from the null value, if the null value really were the mean of the population from which the sample was drawn.  

*$t$-distribution
When you look at a $t$-table (say, in the back of a statistics book), what you are actually looking at is a table of critical values.  That is, the value that the observed $t$ statistic must be greater than in order for the test to be 'significant' at that alpha.  (Typically, these are listed for a small number of possible alphas: $\alpha=\{.10,\ .05,\ .01,\ .001\}$.)  I suspect if you look closely at such tables, they are actually thinking in terms of the degrees of freedom associated with the $t$ statistic in question.  Note that the degrees of freedom for the $t$-statistic is a function of $n$, being $df = n-2$ for a two group $t$-test, or $df = n-1$ for a one group $t$-test (your example seems to be the latter).  This has to do with the fact that the $t$-distribution will converge to a standard normal distribution as the degrees of freedom approaches infinity.  
The way to understand this conceptually is to think about why you need to use the $t$-distribution in the first place.  You know what the reference mean value is that you are interested in and the sample mean that you observed.  If the population from which the samples were drawn was normally distributed (which people are often implicitly assuming), then we know that the sampling distribution of the mean will be normally distributed as well.  So why bother with the $t$-distribution?  The answer is that are not sure what the standard deviation of the population is.  (If we were sure, we really would use the normal distribution, i.e., the $z$-test instead of the $t$-test.)  So we use our sample standard deviation, $\hat\sigma$, as a proxy for the unknown population value.  However, the more data we have, the more sure we can be that $\hat\sigma$ is in fact approximately the right value.  As $n$ approaches the population size (and/or infinity), we can be sure that $\hat\sigma$ in fact is exactly the right value.  Thus, the $t$-distribution becomes the normal distribution.  
A: Well, the short answer is that's what falls out of the math. The long answer would be to do the math$^3$. Instead I'll try to rephrase gung's explanation that these are two different (though related) things. 
You've collected a sample $X_1...X_n$ that is normally distributed with unknown variance$^4$ and want to know if its average is different from some specified value $\mu$. The way you do this is to compute a value that represents how "different" your observations are from the assumption that $\bar{x}=\mu$. Thus the formula for the $t$-statistic$^1$ you presented. Probably the most intuitive way of thinking about why this increases with $n$ is that you have more "confidence" that things are different when you have more samples.
Moving on, this value follows a $t$-distribution$^2$ with $n-1$ degrees of freedom. The way to think about this is that the $t$-distribution is slightly different depending on your sample size. You can see plots of this distribution with 2, 3, 5, and 20 df below.

You'll notice that higher df has more mass in the center and less in the tails of the distribution (I have no intuitive reasoning for why the distributions behave this way, sorry). The critical $t$-value is the x-location where the area under the curve equals a somewhat arbitrary value of your choosing (traditionally 0.05). These values are marked on the graph as points. So for the green curve (df=5), the area under the curve to the left of the left green dot = 0.025, and the area under the curve to the right of the right green dot = 0.025, for a total of 0.05. 
This is why the critical $t$-values decrease with increasing degrees of freedom - as df increases, the critical values must get closer to zero to keep the same area under the curve. And as gung mentioned, as df goes to $\infty$, the curve and critical values will approach that of a standard normal distribution.
So now you have your critical value and your $t$-statistic, and can perform the $t$-test. If your $t$-statistic is greater than the critical value, you then can make the statement that if $\bar{x}=\mu$ really was true, then you would have observed your sample less than 5% (or whatever arbitrary percentage you chose to calculate the critical value for) of the time.

$^1$ Why do we calculate this particular value out of the many arbitrary values we could calculate? Well, this is what falls out of a calculation of a likelihood ratio test$^3$.
If you knew the variance of the samples beforehand, the $z$-statistic (following a normal distribution) mentioned by gung would fall out of this calculation instead, and you would perform a $z$-test
$^2$ Again, this is what falls out of the math$^3$
$^3$ First good result from google: http://math.arizona.edu/~jwatkins/ttest.pdf
$^4$ It turns out the t-test works even if that assumption is not met, but that's a digression
