[Edited to include the connection with Euler's theorem that
there's no nonconstant $4$-term arithmetic progression of squares; briefly,
since $1$ and $4(N^2+N)+1 = (2N+1)^2$ are always squares, $(1,b,c,(2N+1)^2)$
is such a $4$-term progression, so $b=c=1$ and $N^2+N=0$ for any rational
solution. It seems that the same deals with all $k \geq 3$.]
The Diophantine equations
$$ B^2 = b = \frac43(N^2+N) + 1 $$
$$ C^2 = c = \frac83(N^2+N) + 1 $$
have no simultaneous solution in integers other than the eight trivial
solutions with $N=0,-1$, $B = \pm 1$, $C = \pm 1$. Such a result
can be tricky to prove in general, but here we are somewhat lucky
in that these are the only rational solutions, and moreover
$N=0$ and $N=-1$ are the only rational numbers for which the product $bc$
is a square. This can be proved using Fermat's "descent" method;
the calculations, though elementary, may be somewhat laborious
to carry out by hand, but happily this is no longer necessary
thanks to existing tables and software.
In general, if $P,Q$ are quadratic polynomials such that $PQ$ has
four distinct factors then the simultaneous Diophantine equations
$B^2 = P(N)$, $C^2 = Q(N)$ define an elliptic curve. A fundamental
1929
theorem of Siegel assures that such a curve has only finitely many
integral points. The proof is "ineffective" and does not in general
give an algorithm guaranteed to determine all solutions. Later
theoretical and computational advances do provide such an algorithm,
which is feasible at least for $P,Q$ with small coefficients;
but the resulting proof is very far from elementary, and it can be
hard to predict in any given case how hard it is to give an elementary proof.
In the present case, though,
the elliptic curve already has finitely many rational points,
as does the "isogenous" curve $A^2 = P(N) Q(N)$. We bring this curve
$$
A^2 = \left( \frac43(N^2+N) + 1 \right) \left( \frac83(N^2+N) + 1 \right)
$$
into standard Weierstrass form in the usual way starting from the
rational point $(N,A) = (0,1)$: the Taylor expansion of $A$ about $N=0$
starts $1 + 2N + O(N^2)$, so for $N\neq 0$ we can write
$$
A = 1 + 2N + rN^2
$$
for some rational $r$, and divide the resulting equation by $N^2$ to get
$$
(9r^2-32)N^2 + (36r-64)N + (18r-32) = 0.
$$
This equation is quadratic in $N$, and thus
has rational roots iff its discriminant w.r.t. $N$ is a square.
That discriminant factors as $-72(r-2)r(9r-16)$; taking $r=-2x/9$
and removing a factor $(8/3)^2$ we find that
$$
y^2 = x (x+8) (x+9) = x^3 + 17 x^2 + 72 x
$$
for some rational $x,y$. This elliptic curve turns out to have
conductor $24$, so it already appears in Tingley's "Antwerp Tables"
of modular elliptic curves of conductor at most $200$;
it turns out to have label 24C
here.
We find that it has rank zero, and only four rational points, which
must be the point at infinity and the three "$2$-torsion points" with $y=0$.
Alternatively, we can input [0,17,0,72,0] to Cremona's program mwrank
to find that the curve has rank zero, and then find its torsion points
with gp. Either way, we finish by retracing our steps
to find that each of $r=0,2,16/9$
corresponds to one of the known solutions with $N=0$ or $N=-1$
(in each case a double root of the quadratic in $N$), so we are done.
[Thanks to Will Jagy for
calling my attention to this question.]
Added later: It turns out that this $2$-descent calculation
long predates the Antwerp tables: it is essentially equivalent to
Euler's proof of his theorem that there is no nonconstant
$4$-term arithmetic progression of squares. (See for instance
Keith
Conrad's exposition, which states that Euler proved the result in 1780,
answering a question "first raised by Fermat in 1640".)
Indeed if $b$ and $c$ are squares then
$$
1, \
\frac43(N^2+N)+1, \
\frac83(N^2+N)+1, \
4(N^2+N)+1
$$
is such a progression (the last term being $(2N+1)^2$,
so it is constant by Euler, whence $b=c=1$ and we are done.
jamaicanworm writes in a comment that for general $k \geq 3$
the problem is whether there exists a positive integer $N$
such that $c_i := \frac{4(k-i)}{k}(N^2+N) + 1$ is a square
for each $i=1,2,\ldots,k-1$. These $c_i$ form an arithmetic progression,
and extending it by one term in each direction again yields the squares
$c_k = 1$ and $c_0 = (2N+1)^2$. Hence we have an arithmetic progression
of $k+1$ squares, and again Euler's theorem shows that even if we allow
$N$ to be rational the only examples are the trivial ones with $N^2+N=0$.
Suppose the positive integers are $a<b<c<d<e$.
Then $2a<a+b<2b<b+c<2c<c+d<2d<d+e<2e$ give nine distinct sums. So these must be the only sums of pairs.
Then for $a+c$ we have $a+b<a+c<b+c$
So $a+c = 2b$
Similar reasoning will show that
$b+d=2c$;
$c+e=2d$;
$a+d=b+c$;
$b+e=c+d$; and
$a+e=2c$.
From all this we obtain: $(c-b)-(b-a)=(a+c-2b)=0$
So $c-b=b-a$.
Similarly we will end up with $e-d=d-c=c-b=b-a$.
So the five numbers are in arithmetic progression. So their sum is divisible by $5$ (the sum will be $5$ times the middle value).
Best Answer
Let the first number in the sequence of positive integers be $m+1$ and the last number be $n$.
Then
$$\dfrac{n(n+1)}{2}-\dfrac{m(m+1)}{2}=2014$$ $$n^2+n-m^2-m=4028$$ $$(n-m)(n+m)+(n-m)=4028$$ $$(n-m)(n+m+1)=4028$$
As $n$ and $m$ are positive integers, we have $n+m+1$ greater than $n-m$.
Now
$$4028=2*2014=4*1007=19*212=38*106=53*76$$
For the sequence to have maximum number of terms, we need $n-m$ to be a factor that is the highest possible while the other factor, $n+m+1$, is still greater than it. Moreover, for $n$ and $m$ to be integers, $n-m$ and $n+m+1$ must have opposite parity.
Thus $$n-m=53$$and $$n+m+1=76$$ On solving, $$n=64, m=11$$ Thus $$2014=12+13+\ldots+64$$ which is a sequence of $53$ terms.