Can ANOVA be significant when none of the pairwise t-tests is? Is it possible for one-way (with $N>2$ groups, or "levels") ANOVA to report a significant difference when none of the $N(N-1)/2$ pairwise t-tests does? 
In this answer @whuber wrote: 

It is well known that a global ANOVA F test can detect a difference of means even in cases where no individual [unadjusted pairwise] t-test of any of the pairs of means will yield a significant result.

so apparently it is possible, but I do not understand how. When does it happen and what the intuition behind such a case would be? Maybe somebody can provide a simple toy example of such a situation?
Some further remarks:


*

*The opposite is clearly possible: overall ANOVA can be non-significant while some of the pairwise t-tests erroneously report significant differences (i.e. those would be false positives).

*My question is about standard, non-adjusted for multiple comparisons t-tests. If adjusted tests are used (like e.g. Tukey's HSD procedure), then it is possible that none of them turns out to be significant even though the overall ANOVA is. This is covered here in several questions, e.g. How can I get a significant overall ANOVA but no significant pairwise differences with Tukey's procedure? and Significant ANOVA interaction but non-significant pairwise comparisons.

*Update. My question originally referred to the usual two-sample pairwise t-tests. However, as @whuber pointed out in the comments, in the ANOVA context, t-tests are usually understood as post hoc contrasts using the ANOVA estimate of the within-group variance, pooled across all groups (which is not what happens in a two-sample t-test). So there are actually two different versions of my question, and the answer to both of them turns out to be positive. See below.
 A: Summary: I believe that this is possible, but very, very unlikely. The difference will be small, and if it happens, it's because an assumption has been violated (such as homoscedasticity of variance).
Here's some code that seeks out such a possibility. Note that it increments the seed by 1 each time it runs, so that the seed is stored (and the search through seeds is systematic).  
stopNow <- FALSE
counter <- 0
while(stopNow == FALSE) {
  counter <- counter + 1
  print(counter)
  set.seed(counter)
  x <- rep(c(0:5), 100)
  y <- rnorm(600) + x * 0.01
  df  <-as.data.frame( cbind(x, y))
  df$x <- as.factor(df$x)
  fit <- (lm(y ~ x, data=df))
  anovaP <- anova(fit)$"Pr(>F)"[[1]]
       minTtestP <- 1
      for(loop1 in c(0:5)){
        for(loop2 in c(0:5)) {
          newTtestP <- t.test(df[x==loop1,]$y, df[x==loop2,]$y)$p.value
      minTtestP <- min(minTtestP, newTtestP )    
      }
   }

  if(minTtestP > 0.05 & anovaP < 0.05) stopNow <- TRUE 
  cat("\nminTtestP = ", minTtestP )
  cat("\nanovaP = ", anovaP )
  cat("\nCounter = ", counter, "\n\n" )
}

Searching for a significant R2 and no non-significant t-tests I have found nothing up to a seed of 18,000. Searching for a lower p-value from R2 than from the t-tests, I get a result at seed = 323, but the difference is very, very small. It's possible that tweaking the parameters (increasing the number of groups?) might help.   The reason that the R2 p-value can be smaller is that when the standard error is calculated for the parameters in the regression, all groups are combined, so the standard error of the difference is potentially smaller than in the t-test.
I wondered if violating heteroscedasticity might help (as it were). It does. If I use
y <- (rnorm(600) + x * 0.01) * x * 5

To generate the y, then I find a suitable result at seed = 1889, where the minimum p-value from the t-tests is 0.061 and the p-value associated with R-squared is 0.046. 
If I vary the group sizes (which increases the effect of violation of heteroscedasticity), by replacing the x sampling with:
x <- sample(c(0:5), 100, replace=TRUE)

I get a significant result at seed = 531, with the minimum t-test p-value at 0.063 and the p-value for R2 at 0.046.
If I stop correcting for heteroscedasticity in the t-test, by using:
newTtestP <- t.test(df[x==loop1,]$y, df[x==loop2,]$y, var.equal = TRUE)$p.value

My conclusion is that this is very unlikely to occur, and the difference is likely to be very small, unless you have violated the homoscedasticity assumption in regression.  Try running your analysis with a robust/sandwich/whatever you want to call it correction.
A: It's entirely possible:


*

*One or more pairwise t-test is signfiicant but the overall F-test isn't

*The overall F-test is significant but none of the pairwise t-test is


The overall F test tests all contrasts simultaneously. As such, it must be less sensitive (less statistical power) to individual contrasts (eg: a pairwise test). The two tests are closely related to each other but they are not reporting exactly the same thing.
As you can see, the textbook recommendation of not doing planned comparisons unless the overall F-test is significant is not always correct. In fact, the recommendation may prevent us from finding significant differences because the overall F test has less power than planned comparisons for testing the specific differences.
A: *

*The smallest p-value of the t-tests depends on the maximum spread of the different group means (so only two means are important).


*The p-value of the ANOVA test depends on the variance of all the group means (so all the means are important).
For example, the following two situations have the same maximal difference between the means, but a different in-between groups variance.  $$\begin{array}{rclrclrclrcl}
\mu_1& =& -1 ,&\mu_2 &=& 0,& \mu_3 &= &1\\
\mu_1& =& -1 ,&\mu_2 &=& 1,& \mu_3 &= &1\\
\end{array}$$
In this example the t tests will have the same minimum p-value (the largest difference between the means is $2$ in both cases), but the ANOVA test will have different p-values (the in between groups variance is different). This indicates how the ANOVA and t-tests make a difference comparison.
A similar situation is described in the question How can I get a significant overall ANOVA but no significant pairwise differences with Tukey's procedure?
In the answer a scatter plot is made with simulations for the two smallest p-values of the pairwise comparisons, and with colour coding the region is shown where ANOVA would have p-values below 0.05 or 0.1
The pairwise comparisons and the ANOVA test reject the same amount of cases, but they do so in different cases.

The extreme case is when half the groups have a mean around a single point $\mu_a$ and another half of around a single point $\mu_b$. This gives a large variance for in-between groups, whereas the spread can still be modest. The ANOVA is significant while the pairwise comparisons are not.
