Analysis of the problem: Given $a>b$ and $c>d$, we can derive $a+c>b+d$. By rearranging, we get $a-b>d-c$, so option A is correct.

From $a>b$, it follows that $a-c>b-c$, so option C is correct.

Since $c>d$, we have $-c<-d$, which implies $a-c<a-d$, so option D is correct.

For option B, let's verify by substituting values: if $a=1$, $b=3$, $c=2$, and $d=3$, then we find $a+d<b+c$. Therefore, option B does not necessarily hold.

Hence, the correct answer is $\boxed{\text{B}}$.

Key point: Properties of inequalities.

Since $a > b$ and $c > d$, it follows that $a-b > 0$ and $d-c d-c$ always holds, so option A is correct;
Furthermore, because $a > b$, adding $-c$ to both sides yields $a-c > b-c$, so option C is correct;
From $c > d$, we get $-c b+c$, $a+d < b+c$, or $a+d = b+c$, hence it does not necessarily hold,
Therefore, the answer is $\boxed{\text{B}}$.