As I noted earlier the piquancy of the problem has to do with one's gut reaction that there is not enough information provided in order to reach a solution. Compounding this is the natural inclination to pick it up by the wrong end: to commence, that is, rather than to conclude with the identity of the LAN administrator's spouse. A deep breath and a moment's thought will (and indeed did, in the initial response by Another_Scott) between them yield the following preliminary elements about our ten dinner guests:
1. The maximum possible number of handshakes, since selves and spouses are excluded, is 8.
2. By the same token the minimum number of possible handshakes under the term set forth is that roundest of all numbers, 0.
3. Accordingly the various diners shook 8, 7, 6, 5, 4, 3, 2, 1 and no hands for a total of nine unique combinations.
4. This means that two of the diners must have shaken the same number of hands. Since everyone the LAN administrator polled (recall that in framing the puzzle we were at pains to specify that he asked each of the other guests, not himself) responded with a different figure, the LAN guy hisself must have shaken the same number of hands as one of the others. It will be useful for you to put this point aside for the moment without, however, forgetting it entirely.
We have established that someone in the party shook eight hands. We will arbitrarily call this individual "A1" and his/her spouse "A2," maintaining this convention for the other four couples (hence "B1," "B2," "C1," "C2," etc). This means that every member of couples B, C, D and E shook at least one hand (that of the worthy and personable A1) but no more than seven. Having accounted for nine diners we are left with just one candidate for the no-handshakes slot, and this must needs be that pariah, A2.
We turn our attention next to the individual (otherwise unknown except that the pool of suspects is now eight rather than ten) who shook seven hands, and we will call this sociable individual B1. Now we know that each member of the three other couples all shook at least two hands (A1 and B1) but no more than six, and that B1's spouse, the shy and awkward B2, is the only possible candidate for the single-handshake spot, this presumably unwelcome contact having been bestowed by the relentlessly friendly A1.
A pattern emerges:
Couple A: 8 paired with 0
Couple B: 7 paired with 1
Couple C: 6 paired with 2
Couple D: 5 paired with 3
Couple E: 4 paired with 4
Since we had previously determined (at this time be good enough to retrieve preliminary conclusion #4 above) that the LAN administrator shared the same number of handshakes with one other member of the party, that other is, QED, his spouse, and the answer is 4. For right-brain types, [link|http://homepage.mac.com/rcareaga/Sites/aus_pix/handshakes.jpg|table here.]