R, 63 59 bytes

s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])

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-4 bytes thanks to Giuseppe

(The explanation contains a spoiler as to how to solve the problem without computing $k$.)

Explanation:
Let $s$ be the list of remainders. Note the constraint that s[1]<2, s[2]<3, s[3]<5 and s[4]<7. There exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:

• s[2]<2 and s[2]!=s[1]


• s[3]<3 and s[3]!=s[2]


• s[4]<5 and s[4]!=s[3]

The code can probably be golfed further.

Haskell, 69 bytes

Based on the Chinese remainder theorem

m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]

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C# (Visual C# Interactive Compiler), 125 bytes

n=>Enumerable.Range(n+1,209).Any(c=>new[]c%2,c%3,c%5,c%7.OrderBy(x=>x).SequenceEqual(new[]n%2,n%3,n%5,n%7.OrderBy(x=>x)))

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Wolfram Language (Mathematica), 67 bytes

!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7

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Perl 6, 64 61 59 bytes

my&f=bag($_ X%2,3,5,7);any(($_+1+ ^209).map(&f))===f $_

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Fixed my bug by changing set to bag :).

Explanation:

my&f=bag($_ X%2,3,5,7);any(($_+1+ ^209).map(&f))===f $_
 my&f=bag($_ X%2,3,5,7); # helper function: f
 $_ ; # takes the implicit argument $_
 X%2,3,5,7 # cross metaoperator with divide: does cross product division
 bag( ) # stores the results in a bag (weighed set, knows how many identical items it contains)
 ($_+1+ ^209) # Generates a range of the argument +1 up to 209+argument+1
 .map(&f) # Apply the helper function for each value in that range
 any( ) # Is any of these bags
 === # Equal to
 f $_ # The helper function result for the given argument?

Alternative version:

my @a=($_+ ^209).map:bag($_ X%2,3,5,7);@a.shift===any @a

Ruby, 54 bytes

->n[2,3,5,7].each_cons(2).any?n%l!=n%h&&n%h<l

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Uses Robin Ryder's clever solution.

R, 72 bytes

n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F

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Jelly, 15 bytes

8ÆR©PḶ+%Ṣ¥€®ċḢ$

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I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.

PHP, 81 78 72 bytes

while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);

A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.

$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php

$ echo 2|php -nF euc.php

$ echo 1999|php -nF euc.php

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Or 73 bytes with 1 or 0 response

while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;

$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0

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Original answer, 133 127 bytes

function($n)while(++$k<210)if(($r=function($n)foreach([2,3,5,7]as$d)$o[]=$n%$d;sort($o);return$o;)($n+$k)==$r($n))return 1;

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Python 3, 69 bytes

lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210

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Hardcoded