Originally Posted by Swammerdami

Let me show how I solve puzzles of this type. (The cross-hatch grids sometimes presented are generally NOT the best approach.) First I get all the characters in view:
Forenames are H/J/K/L/M/N/O
Surnames are A/B/C/D/E/F/G
Items are knife/mayan/doll/watch/coin/photo/gun
Prices are \$25/30/40/45/50/60/80
(Notice that I replaced "pistol" with "gun" just so I an use "p" as an unambigious abbreviation should that prove convenient. Similarly, I let "m" define the Mayan wagon instead of the ambiguous "w.")

Then I convert each Clue to a List of different people/purchase/price combinations. "Mr." and "Miss" imply possible forenames.

Clue 1
- - mayan s+20
H/J/K/L Elwell - s

Clue 2
M/O Brown - t+20
Hank - - t
Nancy - - u+20
M/O Dolliver - u

Clue 3
- - watch 80/50
- - coin 60/45 = Larry+15
H/J/K/L Fairbanks - 60/45 = photo+15
Nancy - 80/50
- - - 25/30/40
- - - 25/30/40
- - - 25/30/40

Clue 4
H/J/L Clifton gun -
- - doll -
Kenneth - - -
M/N/O Gifford - -

Some of the facts in the Clues won't lend themselves to easy incorporation into these tables. Make a record of them. In this case there was only one:
• Mary spent more than Olivia. (From Clue 2)

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Now the fun begins! Merge information from the separate tables. (Right away we see from Clues 2 and 4 that Nancy's surname must be Gifford. You can also deduce Nancy's purchase by elimination.) But I won't pursue this further, in case someone else wants to give it a whack!
Continuing on:
• In #3, Fairbanks cannot be \$60. (Both photo and coin would be \$45 but no price is duplicated.) So he is \$45; photo is \$30; coin is \$60; Larry is \$45.
• Nancy Gifford spent \$20 more than Dolliver (Clue 2), so not \$20 more than Elwewll (no price is duplicated) so isn't mayan (Clue 1) so is knife by elimination.
• By now, we've got most of the gift/price pairings in the Clue 3 box. We can see that mayan (\$20 more than Elwell) can only be \$45.

Here are the Clue tables as they stand now; information from table #1 has been merged into #3.

Clue 2
M/O Brown - t+20
Hank - - t
Nancy Gifford knife u+20
M/O Dolliver - u

Clue 3 (and 1)
- - watch 80/50
- - coin 60
Larry Fairbanks mayan 45
Nancy Gifford knife 80/50
- - photo 30
H/J/K Elwell - 25
- - - 40

Clue 4
H/J Clifton gun -
- - doll -
Kenneth - - -
Nancy Gifford knife -

• Mary spent more than Olivia. (From Clue 2)

We see that the Clue 3 and Clue 4 tables can be merged. Don't forget to eliminate K (Kenneth, not doll) from Elwell during this merge:

Clue 3 (and 1 and 4)
- - watch 80/50
- - coin 60
Larry Fairbanks mayan 45
Nancy Gifford knife 80/50
- - photo 30
H/J (not K) Elwell doll 25
H/J Clifton gun 40

• Hank is \$20 less than Brown (Clue 2) so cannot be Elwell (he is \$20 less than Fairbanks). Elwell is John; Clifton is Hank by elimination.
• Brown is \$20 more than Hank (Clue 2) so is \$60.
• Nancy Gifford is \$20 more than Dolliver (Clue 2) rather than Brown, so isn't \$80.

By now, the information has been gathered into one odd fact
• Mary spent more than Olivia.

and a single table:
All Clues
- - watch 80
M/O (Mary) Brown coin 60
Larry Fairbanks mayan 45
Nancy Gifford knife 50
M/O (Olivia) Dolliver photo 30
John Elwell doll 25
Hank Clifton gun 40

The leftover names (Kenneth Ainsworth) complete the first row.
Q.E.D.