I definitely think the Putnam exam tests mathematical "maturity" just as much as talent or raw knowledge. By maturity, I mean the somewhat less tangible things we pick up like gauging a problem's difficulty, anticipating the necessary techniques, knowing how to construct an intelligible argument, and knowing when to move on. Ego can be a huge issue as well - I know it was for me. I missed a lot of points on my second try because I tried to solve too many of the problems and ended up with many incomplete solutions instead of a couple complete ones.
I took the exam twice - once as a sophomore barely out of multivariable calculus, and again as a senior. While I had certainly learned more "math" in those two years, I think the real difference was that I had learned how to make a valid mathematical argument. So my 300-level intro to proof writing and analysis class was much more important than my 400-level differential equations and probability classes. I think the only thing that prevented me from doing extremely well was my ego (and a bit of experience).
As far as practicing/studying/preparing, my department offered a "prep" class which was taught by a former IMO medalist. We worked problems, mostly, but he was able to offer a lot of good strategies and tricks as well. The biggest takeaway message seemed to be that preparing to solve a problem is just as important as actually solving it! I.e. taking time to decompose the problem into its core pieces, look for tricks or "dimension reductions," try to figure out what the question is "really" about (i.e. maybe its not a linear algebra problem, but a counting problem). While you can sometimes make progress by jumping in the deep end, a little big-picture can go a long way. This is in stark contrast to the math GRE!
I also can't recommend enough cross-training in the form of reading about problem solving. Books like "How to Solve It", "Proofs from The Book", and blog posts by Terry Tao, etc. are a great resource. Books in a similar vein from the "other" sciences, like Feynman, can be surprisingly helpful as well - the techniques of discovery.
There is no peculiar year in the 21st century, since the middle two digits form a 1 digit number.
In the 20th century, any year is of the form $19xy$. Then the year is peculiar if and only if
$$19+xy=9x \Leftrightarrow 19+10x+y=90+x \Leftrightarrow 9x+y=71 \,.$$
Then since $0 \leq y \leq 9$ we get
$$9x \leq 71 \leq 9x+9 \Rightarrow 62 \leq 9x \leq 71 \Rightarrow x=7 \,.$$
Plugging $x=7$ in $9x+y=71$ we get that $y=8$.
Thus, the only peculiar year in 20's century is 1978.
If the question asks for the peculiar year before 1978, it must be in the 19th century or before:
Repeating:
In the 19th century, any year is of the form $18xy$. Then the year is peculiar if and only if
$$18+xy=8x \Leftrightarrow 18+10x+y=80+x \Leftrightarrow 9x+y=62 \,.$$
Then
$$9x \leq 62 \leq 9x+9 \Rightarrow 53 \leq 9x \leq 62 \Rightarrow x=6 \,.$$
Plugging $x=6$ in $9x+y=62$ we get that $y=8$.
Thus in the 19th century the only peculiar year is $1868$.
P.S. Any peculiar year of the form $1abc$ satisfies
$$10+a+10b+c=10a+b \Rightarrow 9a-9b-c=10$$
This implies that $c \equiv -1 \pmod{9}$ thus $c=8$ and then
$$a-b=2 \,.$$
From here you get easily all the peculiar years between $1000$ and $1999$.
P.P.S. If you are looking for all $4$ digits answers $abcd$ then you need to solve
$$10a+b+10c+d=10b+c \Rightarrow 10a+d= 9(b-c) \,.$$
Then
$$a+d \equiv 0 \pmod 9 \,,$$
and any pair $(a,d)$ which satisfies this relation uniquely determine $b-c$.
So fixing $a$ you get the value(s) of $d$ and from here $b-c$.
Best Answer
When I am working on my research, or on MO/math.SE questions, I often find myself thinking in a way which reminds me of the feel of solving Olympiad problems. If I then solve the problem, I try to find a very special case of my problem which is still challenging, and can be stated and solved using only material on the Olympiad curriculum. I then e-mail Kiran Kedlaya and say "Hey Kiran, do you think this would be a good Olympiad problem?" If he thinks so, he proposes it to the USAMO committe.
I wrote Problem 2 on the 2010 USAMO in this way; it is Theorem 3.2 of this paper specialized to the case that $W_0$ is the group $S_n$. The fact that the "total number of moves" referred to in the theorem is at most $\binom{n}{3}$ is computed in Section 5.2.
I think I send about one problem a year; but most of them get rejected.
UPDATE 2014 Problem B4 of the 2014 Putnam was mine. Let $F(x,y,z) = \sum F_{ijk} x^i y^j z^k$ be a homogenous polynomial of degree $n$ with positive real coefficients. We say that $F$ is hyperbolic with respect to the positive orthant if, for all $(u_1,v_1,w_1)$ and $(u_2,v_2,w_2) \in \mathbb{R}_{> 0}^3$, the polynomial $f(t) = F(tu_1+u_2,tv_1+v_2,tw_1+w_2)$ has $n$ negative real roots.
In this paper, I show that there are constants $V_1$ and $V_2$ (dependent on $n$) so that,
(1) if $F$ is hyperbolic with respect to the positive orthant, then $F_{i(j+1)(k+1)} F_{(i+1)j(k+1)} > V_1 F_{i(j+1)(k+1)} F_{(i+2)jk}$ and the same for all permutations of the indices
(2) if $F_{i(j+1)(k+1)} F_{(i+1)j(k+1)} > V_2 F_{i(j+1)(k+1)} F_{(i+2)jk}$ and the same for all permutations of the indices, then $F$ is hyperbolic with respect to the positive orthant.
The proof is nonconstructive; I also (Theorem 20) give an explicit value of $V_1$. I was thinking about whether I could give a concrete value for $V_2$. The problem was too hard, so I thought instead about homogenous polynomials in two variables, which is the same as inhomogenous polynomials in one variable. At this point, I was basically looking for a converse to Newton's inequality: I wanted a constant $C$ so that, if $a_k^2 > C a_{k-1} a_{k+1}$, then all the roots of $\sum_{k=0}^n a_k z^k$ are real. The result in one variable wasn't worth publishing, but I figured I could make a nice problem by choosing a particular polynomial and asking people to prove the roots were real.
UPDATE 2020 Problem 6 of the 2020 USOMO was mine. It is the key computation from this MO answer, specialized to the case that the matrix $X$ has rank one, so $X_{ij} = x_i y_j$. The rank one case turned out not to be easier than the general case, which is why it doesn't show up in that MO thread, but it is one of the things I thought about when working on that answer, and I noticed at the time that it looked like a strengthening of the rearrangement inequality.