You must have a mistake somewhere because your analytic solutions satisfy the system only when C=0 or J=0, i.e. the trivial case.

pdesys := {diff(C(x,t),t)-phi*epsilon*J(x,t)*C(x,t), diff(J(x,t),x)-epsilon*J(x,t)*C(x,t)}:
Jsol := (x,t)-> J__0*exp(J__0*epsilon*phi*t)/(exp(J__0*epsilon*phi*t)+exp(C__0*epsilon*phi*x)-1):
Csol := (x,t) -> C__0*exp(C__0^epsilon*phi*x)/(exp(J__0*epsilon*phi*t)+exp(C__0*epsilon*phi*x)-1):
subs({J=Jsol,C=Csol},  pdesys): simplify(%);

X := -0.6356560300e-1*ln(Y+200.+17.54410643*Y^(2/3)+102.5985568*Y^(1/3))/a+
         0.6356560300e-1*ln(Y+200.)/a-.2201977080*arctan(.1974510146*Y^(1/3)
         -.5773502693)/a+.1096187623/a:
vx := <1500, 1340.00, 1135.00, 982.00, 884.15, 704.72, 520.00, 287.00, 70.00, 0.>:
vy := <0., 4.28, 7.77, 7.30, 9.00, 13.00, 25.85, 28.91, 38.48, 50.00>:

Statistics:-Fit(X, vy, vx, Y,output=parametervalues);
    [a = 0.115891288738057e-3]

f:=Statistics:-Fit(X, vy, vx, Y):
p1:=plot(vx,vy,style=point):  p2:=plot( [f,Y,Y=0..50],color=red):
plots:-display(p1,p2);

In Maple 2016.1 (64 bit)  I obtain the correct answer:

[12, [x = 1, y = 2]]

This is normal because verify does a kind of syntactic check (extended by some options).

I am a bit surprised that the probabilistic check testeq also fails.

It would have been nice if j:=16117;N:=3;  were provided in the code.
Even nicer if the unknowns had reasonable names (for an outsider).
So, with a simple subs your system is:

sysX := [x6,
 x20-x23,
 x21-0.3505865589e-5,
 x22-0.5304364281e-5,
 x5-x14,
 x24*(-0.3915554290e-1*x12-0.1903748329e-1*x6+0.8260795999e-1)-3.876387504,
 x24*(-0.3860115660e-1*x12-0.1876793978e-1*x6+0.7836678184e-1)-2.040147478*10^6*x23,
 x24*(-0.1876794098e-1*x12-0.9892449327e-2*x6+0.3810204607e-1)-2.040147478*10^6*x9,
 x25*(-0.3915554290e-1*x13-0.1903748329e-1*x7+0.8260795999e-1)-3.876387504,
 x26*(-0.3915554290e-1*x14-0.1903748329e-1*x8+0.8260795999e-1)-3.876387504,
 x1-.9724029753*x6-x12,
 x10-.9724029753*x7-x13,
 x11-.9724029753*x8-x14,
 x3-x12+x10,
 x3-.25*x18-.25*x16,
 x4-x13+x11,
 x25*(-0.3860115660e-1*x13-0.1876793978e-1*x7+0.7836678184e-1)+2.040147478*10^6*x20-7.152482840,
 x25*(-0.1876794098e-1*x13-0.9892449327e-2*x7+0.3810204607e-1)+1.983845478*10^6*x20-5.221405977,
 x26*(-0.3860115660e-1*x14-0.1876793978e-1*x8+0.7836678184e-1)+2.040147478*10^6*x21-10.82168541,
 x26*(-0.1876794098e-1*x14-0.9892449327e-2*x8+0.3810204607e-1)+1.983845478*10^6*x21-8.751240594,
 x15-0.7006679273e-1*x12-.2484955248*x6-1.002451672,
 x16-0.7006679273e-1*x13-.2484955248*x7-1.002451672,
 x17-0.7006679273e-1*x14-.2484955248*x8-1.002451672,
 x18-0.7006679273e-1*x12+.1803623678*x6-1.002451672,
 x19-0.7006679273e-1*x13+.1803623678*x7-1.002451672,
 x2-0.7006679273e-1*x14+.1803623678*x8-1.002451672]:

To solve it with fsolve, Digits must be increased (the system seems to be ill-conditioned).

Digits:=20:
fsolve(sysX,indets(sysX));
{x1 = 1.0112518604279113366, x10 = .50079130545073703755, x11 = -0.42400685353282833597e-1, x12 = 1.0112518604279113366, x13 = .87282458350900007716, x14 = .26867164104671288780, x15 = 1.0733068465024293527, x16 = .96853537340626784353, x17 = .94178275674200828459, x18 = 1.0733068465024293527, x19 = 1.1326128306517202960, x2 = 1.0789746676418324408, x20 = 0.17374637467959646823e-5, x21 = 0.35058655890000000000e-5, x22 = 0.53043642810000000000e-5, x23 = 0.17374637467959646823e-5, x24 = 90.123721949337849443, x25 = 69.574517142133632373, x26 = 49.584072102670758903, x3 = .51046055497717429906, x4 = .91522526886228291076, x5 = .26867164104671288780, x6 = 0., x7 = -.38259166982030833324, x8 = -.31990063204406129237, x9 = 8.4475741104335797627*10^(-7)}

Note that being "almost" linear, using solve instead of fsolve is not a bad idea.

Q:=proc(a,b,c,d,x,y) local t;
   x^2-x+y^2-5*y - evalf(Int(5*exp(sin(t+x)),t=0..b)) end:

Optimization:-NLPSolve('Q(1, Pi/2, 5, 2, x,y)', x= 0 .. 5, y = 0 .. 4);
       [-25.8258678143110600, [x = .750451348084492, y = 2.50000000432642]]

In Maple 2016,

N  := 2:
F  := add(P||k * x^k, k=0..N):
G := [seq(P||k=-1.0 .. 1.0, k=0..N)]:
Explore(plot(F, x=0..1), parameters=G);

works as expected. What version do you use?

Also,

F := a*x:
Explore(plot(F, x=0..1), parameters = [a=0..1]);

works, after replacing (of course) F = a*x with F:= a*x.

But note that you should use [a=0.0 .. 1.0]  otherwise a will take only the two integer values 0 and 1.

In principle,

hyp:=a>0,b>0,c>0,d>0,e>0,f>0,g>0,a>c,a>e,a>g:
is(expr>0) assuming hyp;

should work. But practically, it works only if expr is simple enough.

Unfortunately,

is(exp(a-1)-exp(g-1)>0) assuming hyp;   # ==> FAIL
is((a-1)^3-(g-1)^3>0) assuming hyp;      # ==> FAIL

is(a+b>e) assuming hyp;  # true    (finally!)

You have two issues here.

1. Your definition
f := unapply(rsolve({f(0) = 0, f(1) = 1, f(n) = f(n-1)+n*f(n-2)}, f(n)), n);
is recursive.

2. Maple cannot find a closed form for f(n).

A numeric summation is possible:

g:=rsolve({f(0) = 0, f(1) = 1, f(n) = f(n-1)+n*f(n-2)}, f(n), 'makeproc');
evalf(Sum('g(n)/factorial(n)', n = 0 .. infinity));
    2.929012921

A:=Matrix(20,2, (i,j)->`if`(j=1,i,(i-10)^2)):
B:=Matrix(20,2, (i,j)->`if`(j=1,i,0.75*(i-10)^2)):

PLOT(CURVES(A,LEGEND("1st"),COLOR(RGB,1,0,0)),
          CURVES(B,LEGEND("2nd"),COLOR(RGB,0,1,0))    );

or

pA:=PLOT(CURVES(A,LEGEND("1st"),COLOR(RGB,1,0,0))):
pB:=PLOT(CURVES(B,LEGEND("2nd"),COLOR(RGB,0,1,0))):
plots:-display(pA,pB);

Implicitplot algorithm needs sign changes in order to determine the points.
Mathematically, in this case, all the points of the surface (x+y+z)^2=0 are singular.

Of course, you can use the equivalent implicitplot3d(x+y+z=0, x=-5..5,y=-5..5,z=-5..5) to obtain the plane.

Each of n dancers placed at random positions on the dance-floor randomly chooses one "friend" and one "enemy".
    At each step, every dancer:
        - moves 0.5% closer to the centre of the floor
        - takes a large step towards his friend
        - and a small step away from his enemy.
    At random intervals the dancers re-choose their friends and enemies.

mu = number of steps between frames
rechoose = probability (%) to rechoose friends and enemies after mu*numframes steps.

n:=1000: numframes:=200: mu:=10: rechoose:=10:
x,y:='LinearAlgebra:-RandomVector(n,generator=-1.0 ..  1.0,datatype=float[8])' $2:
FE:=proc() global F,E; #friend,enemy
  F,E:='LinearAlgebra:-RandomVector(n,generator=1..n,datatype=integer[4])' $2;
end:
FE():
step:=proc(x::Vector(datatype=float[8]),y::Vector(datatype=float[8]),
      E::Vector(datatype=integer[4]),F::Vector(datatype=integer[4]))
option autocompile;
local i::integer[4],ex::float[8],ey::float[8],ed::float[8],
                    fx::float[8],fy::float[8],fd::float[8];
to mu do
for i to n do
  ex:=x[E[i]]-x[i];ey:=y[E[i]]-y[i]; ed:=sqrt(ex^2+ey^2);
  fx:=x[F[i]]-x[i];fy:=y[F[i]]-y[i]; fd:=sqrt(fx^2+fy^2);
  x[i] := 0.995*x[i] - 0.01*ex/(ed+0.01) + 0.02*fx/(fd+0.01);
  y[i] := 0.995*y[i] - 0.01*ey/(ed+0.01) + 0.02*fy/(fd+0.01);
od; od;
end:
for k to numframes do
  p[k]:=plot(x,y);
  step(x,y,E,F);
  if rand(1..100)()<=rechoose then FE() fi;
od:
plots[display](seq(p[k],k=1..numframes),style=point,axes=none,insequence=true);


Edited. Added definition, comments and a few changes in code.

Your system is incompatible. You have posted a similar system recently and you were told how to see this (e.g. remove the last equation).

allcomp:=proc(F, n)
local M:=Iterator:-CartesianProduct(F$n);
[seq(`@`(entries(u,nolist)), u=M)];
end;

allcomp([sin, cos, t->t^2], 3)(x);  #example

For Maple<2016 use:
allcomp2015:=proc(F,n)
local T:=combinat:-cartprod([F$n]),r:=NULL;
while not T[finished] do r:=r,`@`(op(T[nextvalue]()))  end do;
[r]
end;

To integrate f(x) over the interval [a,b] you can use

Int(f(x)*Heaviside(x-a)*Heaviside(b-x), x=-infinity..infinity);

But why would you want to do so, when Maple is able to find the antiderivative?

J:=int(c-sqrt(a+b*(v*x+u)^2), x);
x1,x2 := -(b*u+sqrt(b*c^2-a*b))/(b*v) , (-b*u+sqrt(b*c^2-a*b))/(b*v);
eval(J,x=x2) - eval(J,x=x1); simplify(%);